Chapter 1. Electric Charges and Fields.NCRT Question and ... 12TH/CLASS 12TH PHYS… · • Solution: The forces acting on charge q at A due to charges at B and -q at C are F12 along

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1. How can you charge a metal sphere positively without touching it?

• Solution: (a) shows an uncharged metallic sphere on an insulating metal stand. Bring a negatively charged rod close to the metallic sphere, as shown in [fig (b)]. As the rod is bought close to the sphere, the free electrons in the sphere move away due to repulsion and start piling up at the farther end. The near end becomes positively charged due to deficit of electrons. This process of charge distribution stops when the net force on the free electrons inside the metal is zero. Connect the sphere to the ground while the positive charges at the near end will remain held there due to the attractive force of the negative charges on the rod, as shown in [fig (c)]. Disconnect the sphere from the ground. The positive charge continues to be held at the near end [fig (d)]. Remove the electrified rod. The positive charge will spread uniformly over the sphere as shown in [fig (e)].

In this experiment, the metal sphere gets charged by the process of induction and the rod does not lose any of its charge. Similar steps are involved in charging a metal sphere negatively by induction, by bringing a positively charged rod near it. In this case the electrons will flow from the ground to the sphere when the sphere is connected to the ground with a wire.

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MATHEMATIC CENTER D96 MUNIRKA VILLAGE NEW DELHI 110067 & VIKAS PURI NEW DELHI CONTACT FOR COACHING MATHEMATICS FOR 11TH 12TH NDA DIPLOMA SSC CAT SAT CPT CONTACT FOR ADMISSION GUIDANCE B.TECH BBA BCA, MCA MBA DIPLOMA AND OTHER COURSES 09810144315

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2. If 109 electrons move out of a body to another body every second, how much time is required to get a total charge of 1C on the other body?

• Solution: In one second 109 electrons move out of the body. Therefore the charge given out in one second

is . The time required to accumulate a charge of 1 C can then be estimated to be

.

In one second 109 electrons move out of the body. Therefore the charge given out in one second

is . The time required to accumulate a charge of 1 C can then be estimated to

be

Thus to collect a charge of one Coulomb, from a body from which 109 electrons move out every second, we will need approximately 200 years. One Coulomb is, therefore, a very large unit for many practical purposes.

It is, however, also important to know what is roughly the number of electrons contained in a piece

of one cubic centimeter of material. A cubic piece of copper of side 1 cm contains about electrons.

Thus to collect a charge of one Coulomb, from a body from which 109 electrons move out every second, we will need approximately 200 years. One Coulomb is, therefore, a very large unit for many practical purposes.

It is, however, also important to know what is roughly the number of electrons contained in a piece

of one cubic centimeter of material. A cubic piece of copper of side 1 cm contains about electrons.

3. How much positive and negative charge is there in a cup of water?

• Solution: Let us assume that the mass of one cup of water is 250 g. the molecular mass of water is 18 g.

Thus, one mole (= molecules) of water is 18 g. Therefore the number of molecules in

one cup of water is (250/18) .Each molecule of water contains two hydrogen atoms and one oxygen atom, i.e., 10 electrons and 10 protons. Hence the total positive and total negative

charge has the same magnitude. It is equal to (250/18)

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4. Coulomb's law for electrostatic force between two point charges and Newton's law for

gravitational force between two stationary point masses, both have inverse-square law dependence on the distance between the charges/masses. (a) Compare the strength of these forces by determining the ratio of their magnitudes

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(i) For and electron and a proton and (ii) for two protons. (b) Estimate the accelerations of electron and proton due to the electrical force of their mutual attraction when they are 1 Å (=10-10 m) apart? (mp = 1.67

).

• Solution: (a) (i) The electric force between an electron and a proton at a distance r apart is:

where the negative sign indicates that the force is attractive. The corresponding gravitational force 9always attractive) is:

where mp and me are the masses of a proton and an electron respectively.

(ii) On similar lines, the ratio of the magnitudes of electric force to the gravitational force between two protons at a distance r apart is:

However, it may be mentioned here that the signs of the two forces are different. For two protons, the gravitational force is attractive in nature and the Coulomb force is repulsive. The actual values of

these forces between two protons inside a nucleus (distance between two protons is m

inside a nucleus) are Fe ~ 230 N whereas FG ~ .

The (dimensionless) ratio of the two forces shows that electrical forces are enormously stronger than the gravitational forces.

(b) The electric force F exerted by a proton on an electron is same in magnitude to the force exerted by an electron on a proton; however the masses of an electron and a proton are different. Thus, the

magnitude of force is

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Using Newton's second law of motion, F= ma, the acceleration that an electron will undergo is a=

Comparing this with the value of acceleration due to gravity, we can conclude that the effect of gravitational field is negligible on the motion of electron and it undergoes very large accelerations.

The value for acceleration is .

5. A charged metallic sphere A is suspended by a nylon thread. Another charged metallic

sphere B held by an insulating handle is brought close to A such that the distance between their centers is 10 cm. The resulting repulsion of A is noted (for example, by shining a beam of light and measuring the deflection of its shadow on a screen). Spheres A and B are touched by uncharged spheres C and D respectively. C and D are then removed and B is brought closer to A to a distance of 5.0 cm between their centers. What is the expected repulsion of A on the basis of Coulomb’s law? Spheres A and C and spheres B and D have identical sizes. Ignore the sizes of A and B in comparison to the separation between their centers.

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• Solution: Let the original charge on the sphere A be q and that on B be q’. At a distance r between their

centers, the magnitude of the electrostatic force on each is given by F=

Neglecting the sizes of spheres A and B in comparison to r. When an identical but uncharged sphere C touches A, the charges redistribute on A and C and, by symmetry, each sphere carries a charge q/2. Similarly, after D touches B, the redistributed charge on each is q’/2. Now, if the separation between A and B is halved, the magnitude of the electrostatic force on each is F’

=

Thus the electrostatic force on A, due to B, remains unaltered.

6. Consider three charges q1, q2, q3 each equal to q at the vertices of an equilateral

triangle of side l. What is the force on a charge Q (with the same sign as q) placed at the centroid of the triangle?

• Solution: In the given equilateral triangle ABC of sides of length l, if we draw a perpendicular AD to the side BC.

AD = AC cos 30o = ( l and the distance AO of the centroid O from A is (2/3) AD =(1/ ) l.

By symmetry AO=BO=CO. Thus, Force F1 on Q due to charge q at A = along AO

Force F2 on Q due to charge q at B = along BO

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Force F3 on Q due to charge q at C = along CO

The resultant of forces F2 and F3 is along OA, by the parallelogram law. Therefore, the

total force on Q = (r^-r^)=0, where r^is the unit vector along OA. It is clear also by symmetry that the three forces will sum to zero. Suppose that the resultant force was non-zero but in some direction.

7. Consider the charges q, q , and –q placed at the vertices of an equilateral triangle.

What is the force on each charge?

• Solution:

The forces acting on charge q at A due to charges at B and -q at C are F12 along BA and F13 along AC respectively. By the parallelogram law, the total force F1 on the charge q at A is given by F1= F r1

^Where r1^ is a unit vector along BC. The force of attraction or repulsion for each pair of charges

has the same magnitude F= . The total force F2 on charge q at B is thus F2 = F r2^ Where

r2^ is a unit vector along AC. Similarly the total force on charge -q at C is F3 = F n^ , where n^

is the unit vector along the direction bisecting the <BCA. It is interesting to see that the sum of the forces on the three charges is zero, i.e.,F1+F2+F3 = 0.

The result is not at all surprising. It follows straight from the fact that Coulomb's law is consistent with Newton's third law.

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8. An electron falls through a distance of 1.5 cm in a uniform electric field of magnitude 2.0 × 104 NC-1. The direction of the field is reversed keeping its magnitude unchanged and a proton falls through the same distance. Compute the time of fall in each case. Contrast the situation with that of 'free fall under gravity'.

• Solution: In the figure, the field is upward, so the negatively charged electron experiences a downward force of magnitude eE where E is the magnitude of the electric field. The acceleration of the electron is Ae = eE/me Where me is the mass of the electron. Starting from rest, the time required by the electron to fall through a distance h is given by

In the next figure, the field is downward, and the positively charged proton experiences a downward

force of magnitude eE. The acceleration of the proton is where mp is the mass of the proton;

The time of fall for the proton is

. Thus, the heavier particle (proton) takes a greater time to fall through the same distance. This is in basic contrast to the situation of 'free fall under gravity' where the time of fall is independent of the mass of the body. Note that in this example we have ignored the acceleration due to gravity in calculating the time of fall. To see of this is justified, let us calculate the acceleration of the proton in the given electric field:

In the figure, the field is upward, so the negatively charged electron experiences a downward force of magnitude eE where E is the magnitude of the electric field. The acceleration of the electron is

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Ae = eE/me Where me is the mass of the electron. Starting from rest, the time required by the electron to fall through a distance h is given by

In the next figure, the field is downward, and the positively charged proton experiences a downward

force of magnitude eE. The acceleration of the proton is where mp is the mass of the proton;

The time of fall for the proton is

. Thus, the heavier particle (proton) takes a greater time to fall through the same distance. This is in basic contrast to the situation of 'free fall under gravity' where the time of fall is independent of the mass of the body. Note that in this example we have ignored the acceleration due to gravity in calculating the time of fall. To see of this is justified, let us calculate the acceleration of the proton in the given electric field:

which is enormous compared to the value of g (9.8 ms-2), the acceleration due to gravity. The acceleration due to gravity can be ignored in this example, which is enormous compared to the value of g (9.8 ms-2), the acceleration due to gravity. The acceleration due to gravity can be ignored in this example.

9. Two point charges q1 and q2, of magnitude +10–8 C and –10–8 C, respectively, are

placed 0.1 m apart. Calculate the electric fields at points A, B and C as shown in figure.

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• Solution: The electric field vector E1A at A due to the positive charge q1 points towards the right and has a

magnitude

The electric field vector E2A at A due to the negative charge q2 points towards the right and has the same magnitude. Hence the magnitude of the total electric field EA at A is

EA is directed toward the right. The electric field vector E1B , at B due to the positive charge q1 points

towards the left and has a magnitude

The electric field vector E2B at B due to the negative charge q2 points towards the right and has a magnitude

The magnitude of each electric field vector at point C, due to charge q1 and q2 is

The directions in which these two vectors point are indicated in the figure. The resultant of these two

vectors is . EC points towards the right.

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10. Two charges ±±±±10µµµµC are placed 5.0 mm apart. Determine the electric field at 9a) A point P on the axis of the dipole 15 cm away from its center O on the side of the positive charge , as shown in the figure (a), and (b) a point Q, 15 cm away from O on a line passing through O and normal to the axis of the dipole, as shown in the figure (b).

• Solution: (a) Field at P due to charge +10µC

= along BP

Field at P due to charge -10 µC

= along PA

The resultant electric field at P due to the two charges at A and B is along BP.

In this example, the ratio OP/OB is quite large (=60). Thus, we can expect to get approximately the same result as above by directly using the formula for electic field at a far-away point on the axis of

a dipole. For a dipole consisting of charges , 2a apart, the electric field at a distance r from the center on the axis of the dipole has a magnitude

E= (r/a>>1)

Where p=2aq is the magnitude of the dipole moment

The direction of electric field on the dipole axis is always along the direction of the dipole moment

vector (i.e. from -q to q). Here, p=

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Therefore,

along the dipole moment direction AB, which is close to the result obtained earlier

o Field at Q due to charge +10µC at B

along QA. Clearly, the components of these two forces with equal magnitudes cancel along the direction OQ but add up along the direction parallel to BA. Therefore, the resultant electric field at Q due to the two charges at A and B is

As in (a), we can expect to get approximately the same result by directly using the formula for dipole field at a point on the normal to the axis of the dipole:

E (r/a>>1)

The direction of electric field in this case is opposite to the direction of the dipole moment vector. Again the result agrees with that obtained before.

11. The electric field components in the figure are in which

====800 N/C m .Calculate (a) the flux through the cube, and (b) the charge within the cube. Assume that a=0.1 m.

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• Solution: (a) Since the electric field has only an x component , for faces perpendicular to x direction, the angle

between E and ∆S is . Therefore, the flux Φ= E. Φ is separately zero for each face of the cube except the two shaded ones. Now the magnitude of the electric field at the left face is

(x=a at the left face). The magnitude of the electric field at the right face is

(x= 2a at the right face). The corresponding fluxes are

Net flux through the cube

(b) We can use Gauss's law to find the total charge q inside the cube. We have φ=q/Є0 ot q= φ Є0 .

Therefore, Q = .

12. An electric field is uniform, and in the positive x direction for positive x, and uniform

with the same magnitude but in the negative x direction for negative x. It is given that E = 200 i^ N/C for x>0 and E=-200 i^ N/C for x<0. A right circular cylinder of length 20 cm and radius 5 cm has its center at x= -10 cm. (a) What is the net outward flux through each flat face?

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(b) What is the flux through the side of the cylinder?

(c) What is the net outward flux through the cylinder?

(d) What is the net charge inside the cylinder?

• Solution:

(a) We can see from the figure that on the left face E and ∆S are parallel. Therefore, the outward

flux is i^.∆S

=+200 ∆S, since i^.∆S=-∆S

=+200 =+1.57 Nm2C-1

(b) On the right face, E and ∆S are parallel and therefore

(c) For any point on the side of the cylinder E is perpendicular to ∆S and hence E. ∆S=0. Therefore, the flux out of the side of the cylinder is zero. Net outward flux through the cylinder Φ=1.57+1.57+0=3.14 Nm2C-1

(d) The net charge within the cylinder can be found by using Gauss's law which gives Q =Є0 Φ

= 3.14

= 2.78 .

13. An early model for an atom considered it to have a positively charged point nucleus

of charge Ze, surrounded by a uniform density of negative charge up to a radius R. The atom as a whole is neutral. For this model, what is the electric field at a distance r from the nuclus?

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• Solution: The charge distribution for this model of the atom is as shown in the figure. The total negative charge in the uniform spherical charge distribution of radius R must be –Ze, since the atom (nucleus of the charge Ze + negative charge) is neutral. This immediately gives us the negative charge density ρ, since we must have

. To find the electric field E ( r) at a point P which is a distance r away from the nucleus, we use Gauss’s law. Because of the spherical symmetry of the charge distribution, the magnitude of the electric field E(r) depends only on the radial distance of r. Its direction is along (or opposite to) the radius vector r form the origin to the point P. The obvious Gaussian surface is a spherical surface centred at the nucleus. We consider two situations, namely r<R and r >R. r<R : The electric flux

enclosed by a spherical surface is

where E(r) is the magnitude of the electric field at r. This is because the field at any point on the spherical Gaussian surface has the same direction as the normal to the surface there, and has the same magnitude at all points on the surface. The charge q enclosed by the Gaussian surface is the

positive nuclear charge and the negative charge within the sphere of radius r. i.e q = Ze +

Substituting for the charge density obtained earlier, we have Q= Ze-Ze

Gauss’s law then gives, E(r)=

The electric field is directed radially outward. r>R: In this case, the total charge enclosed by the

Gaussian spherical surface is zero since the atom is neutral. Thus, from Gauss’s law, E(r) or E(r) = 0 ;r>R

At r = R, both cases give the same result : E=0

14. What is the force between two small charged spheres having charges of 2 × 10-7 C

and 3 × 10-7 C placed 30 cm apart in air?

• Solution: The charges of the I sphere q1 = 2 × 10-7 C The charges of the II sphere q2 = 3 × 10-7 C The distance between them r = 30 cm The constant k = 9 × 109 Nm2/C2

Therefore, the force between them =

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= = 6 ×10-3 N (repulsive).

15. The electrostatic force on a small sphere of charge 0.4 C in air due to another small

sphere of charge -0.8 C in air is 0.2 N. (a) What is the distance between the two spheres? (b) What is the force on the second sphere due to the first?

• Solution: (a) The charge on I sphere = 0.4 C The charge on II sphere = -0.8 C The force between the spheres = 0.2 N The constant k = 9 109 Nm2/C2

Therefore, the distance between them = r =

=

= = 0.12 m or 12 cm (b) The force on the II sphere due to the first is also 0.2N and is an attractive force.

16. Check that the ratio ke2/G me mp is dimensionless. Look up a table of physical

constants and determine the value of this ratio. What does the ratio signify?

• Solution: The constant 'k' = 9 × 109 Nm2/C2 The charge 'e' = 1.602 × 10-19 C Gravitational constant 'G' = 6.67 × 10-11 m3/s2 kg The mass of the electron 'me' = 9.11 × 10-31 kg The mass of the proton 'mp' = 1.67 ×10-27 kg

The dimension of 'k' = Nm2/C2 = The dimension of 'e2 ' = C2

The dimension of 'G' =

Therefore = = = Dimensionless

The value of = 2.276 × 1039. The force signifies the ratio between the electrical and the gravitational force between two particles, an electron and a proton. It also shows that the electrical force is much greater than the gravitational force between an electron and a proton.

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17. (a) Explain the meaning of the statement "electric charge of a body is quantized".

(b) Why can one ignore quantization of electric charge when dealing with macroscopic i.e., large-scale charges?

• Solution: (a) The magnitude of the smallest amount of charge is e = 1.602 x 10-19 C. Any charged body, large or small, has a total charge which is an integral multiple of e, namely, ±€e, ±€2e, ±€3e………no fractions of +e have been seen. Hence, electric charge of a body is quantized. (b) For macroscopic charges for which Q = ne, where n is a very large number, quantization of charge can be ignored. It is similar to supposing that water is a continuous fluid - for most practical purposes we can ignore the fact that actually it consists of individual molecules.

18. When a glass rod is rubbed with a silk cloth, charges appear on both the sides. A

similar phenomenon is observed with many other pairs of bodies. Explain how this observation is consistent with the law of conservation of charge?

• Solution: When a glass rod gets positively charged by rubbing with a silk cloth, in the process the silk gets negatively charged. This can be checked as follows: If the cloth is taken near the charged glass rod, hanging from a thread, the rod is attracted by and swings towards the cloth. Therefore the rubbing of the rod by the cloth results in both becoming charged, one positively and the other negatively. The total charge remains zero, as it was to begin with. This suggests that what happens in frictional, electricity is a transfer of electric charge from one body to the other and not creation of charge. Thus if some amount of charged matter is contained in a certain volume of space and no charged matter is allowed to enter or leave this volume, then the total charge within the volume will never change. This explains the law of conservation of charge, which states that the total charge of an isolated system does not change.

19. Four point charges qA = 2 µC, qB = -5µC , qC = 2µC are located at the corners of

a square ABCD of side 10 cm. What is the force on the charge of 1 µ C placed at the centre of the square?

• Solution:

Consider a unit positive charge being placed at P, the center of the square. Let the distance between the center P and any of the four charges at A, B, C and D

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PA = PB = PC = PD = r (say) The force due to the charge at A (+2) is repulsive and is directed along PC F1 = k(2)/r2

The force due to the charge at B (-5 ) is attractive and is directed along PB F2 = k(5)/r2 The force due to the charge at C (2) is repulsive and is directed along PA F3 = k(2)/r2

The force due to the charge at D (-5) is attractive and is directed along PD F4 = k(2)/r2

The net force at P is F = F1 + F2 + F3 + F4

The forces F1 and F3 are equal in magnitude and are opposite in direction at P Thus F1 = -F3

Similarly, the forces F2 and F4 are equal in magnitude and are opposite in direction and. Thus F2 = -F4

Therefore the net force at P, F = 0.

20. (a) An electrostatic line of force is a continuous curve. That is, a line of force cannot

have sudden breaks. Why not?

(b) Explain why two lines of force never cross each other at any point.

• Solution: (a) A line of force is a line drawn in such a way that the tangent at each point on the line gives the direction of electric field at that point. In general the direction of the electric field vector changes from point to point. Therefore, lines of force are, generally, curved lines. In the special case of a field, which is uniform in a region of space, the lines of force become parallel straight lines.

(b) To construct a line of force, take a point p in space and let the electric field vector there be .

Move from p by a very small distance in the direction of , reaching a nearby point p′ . Let

the electric field at p′ be . Then move from p′ in the direction of ′ . To a nearby point p′ ′ . And so

on from p′ to p′ ′ , in the direction of ′ ′ ′ etc. By bringing these points closer and closer, a

continuous curve starting at p in the direction of is got. Thus as the field is continuous the line of force is also continuous. If we suppose that the two lines of force cross each other, then it would be possible to draw two tangents to the two lines of force at the point of intersection. This would directly mean two different directions of electric field at a single point. This would further imply a physically impossible situation in which a unit positive charge placed at the intersection would simultaneously move along two different paths. Hence, intersection of two lines of force is not possible.

21. Two point charges qA = 3µC and qB = - 3µC are located 20 cm apart in vacuum.

(a) What is the electric field at the midpoint O of the line AB joining two charges?

(b) If a negative test charge of magnitude 1.5 × 10-9C is placed at this point, what is the force experienced by the net charge?

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• Solution: (a) The point charge qA = 3µC The point charge qB = - 3µC Let the mid point be O between the two point charges. The distance O from qA = 10 cm The distance O from qB = 10 cm

The electric field E1 at 0 due to qA =

= = 27 × 105 NC-1 is directed from A to 0.

The electric field at O due to qB = = 27 × 105 NC-1 is directed from O to B . Therefore, the total electric field at O = 54 × 105 NC-1. The field is directed towards qB. (b) If a negative charge of magnitude 1.5 × 10-9 is placed at O The force is given by the formula F = E × q where E is the electrical field intensity at O and q is the magnitude of the charge. Thus the force at the midpoint O is equal to F = 54 × 105 × 1.5 10-9 = 81 × 10-4 N and is directed from B to A since the force due to B on a negative charge is repulsive and the force due to a positive charge at a is attractive. Thus the net force is from B to A.

22. A system has two charges qA = 2.5 ×10

-7 C and qB = -2.5 × 10-7 C located at points A:

(0, 0, -15 cm) and B: (0, 0, +15 cm), respectively. What are the total charge and electric dipole moment of the system?

• Solution:

The points A and B lie on Z-axis. From the above fig, AB = OA + OB = 15 + 15 = 30 cm = 0.3 m Total charge q = qA + qB

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= 2.5 × 10-7 + (-2.5 ×10-7) = 0 Electric dipole moment p = (qA or qB) × AB = 2.5 ×10-7 × 0.3 = 7.5 × 10-8 cm.

23. An electric dipole moment 4 × 10-9 cm is aligned at 30° with the direction of a

uniform electric field of magnitude 5 × 104 NC-1. Calculate the magnitude of the torque acting on the dipole.

• Solution: Electric dipole moment p = 4 × 10-9 cm The uniform electric field E =5 × 104 NC-1 The angle of dipole to the field = 30°

Therefore, the torque on the dipole = = 4 × 10-9 × 5 × 104 × sin30o = 10 × 10-5 = 1 × 10-4 Nm.

24. A polythene piece rubbed with wool is found to have a negative charge of 3 × 10-7 C.

(a) Estimate the number of electrons transferred (from which to which)?

(b) Is there a transfer of mass from wool to polythene?

• Solution: (a) The negative charge on polythene = 3 × 10-7 C Number of electrons in 1 C = 6.25 × 1018 Therefore, number of electrons in polythene = 3 × 10-7 × 6.25 × 1018 = 1.875 × 1012 electrons. The electrons have transferred from wool to polythene. (b) The mass transferred = Number of electrons × Mass of electron = 1.875 × 1012 × 9.11 10-31

= 1.7 × 10 -18 kg Therefore the transfer of mass from wool to polythene is very small.

25. (a) Two insulated charged copper spheres A and B have their centres separated by a

distance 50 cm. What is the mutual force of electrostatic repulsion if the charge on each is 6.5 × 10-7 C? The radii of A and B are negligible compared to the distance of separations. (b) What is the force of repulsion if each sphere is charged double the above amount and the distance between them is halved?

• Solution: (a) The value of charge on each sphere q1 = q2 = 6.5 × 10-7 C

The distance between them 'r' = 50 cm. Therefore the force of repulsion 'F' =

= = 1.5 × 10-2 N. (b) The value of charge is doubled = (6.5 × 10-7 ) 2C

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The distance between them halved 'r' = 25 cm

Therefore, the force of repulsion F = = 0.24 N.

26. Figure below shows tracks of three charged particles in a uniform electrostatic field.

Give the signs of the three charges. Which particle has the highest charge to mass ratio?

• Solution:

Particle 1 is deflected towards the positive plate; hence it must have a negative charge. Particle 2 deflects towards the positive plate so it must be negatively charged. Particle 3 is deflected towards the negative plate, hence it must have a positive charge. Suppose the particles take time 't' to travel a distance CD between the plates, i.e. they are under the influence of the electric field for time t.

Then, the deflection of the particle is given by y = t2. where e/m is the charge to mass ratio of the particle. It is clear that y is proportional to e/m. Since the particle 3 has a higher deflection, it must have a higher charge to mass ratio.

27. Consider a uniform electric field E = 3 × 103 î N/C

(a) What is the flux of this field through a square of 10cm on a side whose plane is parallel to the yz plane?

(b) What is the flux through the same square if the normal to its plane makes a 60° angle with the x axis?

• Solution: The electric field 'E' = 3 × 103 î N/C Side of the square = 10cm = 0.10m

... The flux of this field φ Ε = = 3 × 103 × 0.10 × 0.10 (î.î) = 30 Nm2 C-1

(b) Normal to the plane makes an angle = 60° ... The flux through the square φ = ES cos 60°

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... φ Ε = = 25.95 Nm2C-1.

28. Careful measurement of the electric field at the surface of a black box indicates that

the net outward flux through the surface of the box is 8.0 ×103 Nm2/C. (a) What is the net charge inside the box?

(b) If the net outward flux through the surface of the box were zero, could you conclude that there were no charges inside the box? Why or why not?

• Solution: Flux through the surface φ = 8.0 × 103 Nm2 /C (a) The net charge inside the box is q = ε 0 φ = 8.85 × 10-12 × 8.0 × 103 = 0.07 µ C. (b) No, we cannot conclude that there was no charges inside the box, if the net downward flux through the surface of the box was zero unless the net charge inside is zero.

29. A point charge +10µµµµ C is a distance 5 cm directly above the centre of a square of side

10cm as shown in the figure. What is the magnitude of the electric flux through the square?

.

• Solution: The given square can be imagined as one of the six faces of a cube of side 0.10m. The given charge can be imagined to be at the centre of this cube.

... The electric flux through the square =

= Nm3 C-1 = 1.88 × 105 Nm2 C-1.

30. A point charge of 2.0 µC is at the centre of a cubic Gaussian surface 9.0 cm on edge.

What is the net electric flux through the surface?

• Solution: Point charge q = 2.0µC = 2 × 10-6 C

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... The net electric flux through the surface is given by

φ = Nm2 C-1 = 2.26 × 105 Nm2C-1.

31. A point charge causes an electric flux of -1.0 × 103 Nm2/C to pass through a spherical Gaussian Surface of 10.0 cm radius centered on the charge. (a) If the radius of the Gaussian Surface were doubled, how much flux would pass through the surface? (b) What is the value of the point charge?

• Solution: Electric flux φ = -1.0 × 103 Nm2/C Radius of the spherical Gaussian surface = 10.0cm Therefore r = 0.1m (a) If the radius of the Gaussian surface is doubled i.e. r = 2r = 2 × 0.1 = 0.2m Electric flux φ = -103 Nm2/C Because the charge enclosed is the same in the two cases. (b) The point charge 'q' is given as q = ε 0φ

= = -8.8 × 10-9C.

32. A conducting sphere of radius 10 cm has an unknown charge. If the electric field 20

cm from the center of the sphere is 1.5 * 103 N/C and points radially inward, what is the net charge on the sphere?

• Solution: Radius of the sphere R = 10cm = 0.1m Electric field E = 1.5 × 103 N/C Distance of the field from the centre 'r' = 20cm = 0.2m

We know that E(r)

... q = E(r) 4π ε0 r2

= 6.67 × 10-9 C = 6.67 nC Since the points are radially inward the charge 'q' is given as -6.67 nC.

33. A uniformly charged conducting sphere of 2.4m diameter has a surface charge

density of 80.0 µC/m2. (a) Find the charge on the sphere.

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(b) What is the total electric flux leaving the surface of the sphere?

• Solution: (a) Diameter of the sphere = 2.4m ... Radius of the sphere 'R' = 1.2m Surface charge density of the sphere = 80.0 µC/m2 We know that the total charge q is given as = 4πR2 σ = 4 × 3.14 × (1.2)2 × 80.0 × 10-6 = 1.45 × 10-3 C. (b) The total electric flux leaving the surface of the sphere is given as

φ = = 1.45 × 10-3 × 4 × 10-9 = 1.6 × 108 Nm2/C.

34. An infinite line charge produces a field of 9 x 104 N/C at a distance of 2cm. Calculate

the linear charge density?

• Solution: Electric field of the infinite line charge 'E' = 9 × 104 N/C Distance of the field 'x' = 2cm = 0.02m

We know that E =

... λ =

λ = = 10-7 cm-1.

35. An oil drop of 12 excess electrons is held stationary under a constant electric field of

2.55 ×××× 104

NC-1 in Millikan's oil drop experiment. The density of the oil is 1.26 g cm-3. Estimate the radius of the drop. (g = 9.81 ms-2 , e = 1.6 x 10-19 C).

• Solution: Number of electrons 'n' = 12 The electric field 'E' = 2.55 × 104 NC-1 The density of oil 'ρ' = 1.26 × 103 kg/m3 The weight of drop of oil = m × g

The mass of oil drop 'm' = The acceleration due to gravity g = 9.81 m/s2 . The balancing weight = n × e × E = 12 × 1.6 × 10-19 ×€2.55 x 104 Therefore, mg = n × e × E

= 12 × 1.6 × 10-19 × 2.55 × 104

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Therefore, r3 = or 'r'= 9.8 × 10-7m.

36. Which among the curves shown in the figures cannot possibly represent electrostatic

field lines?

(c) (d)

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(e)

• Solution: Only (c) represents the correct field lines, whereas the other cases are not possible because of the following reasons: (a) Electric field lines do not originate normal to the surface. (b) Electric lines of force do not start from a negatively charged point. (d) Electric lines of force cannot intersect each other. (e) Electric lines of force cannot form closed loops.

37. In a certain region of space, electric field is along the z-direction through out. The

magnitude of electric field is, however, not constant but increases uniformly along the positive z-direction, at the rate of 105 NC-1 per metre. What are the forces and torque experienced by a system having a total dipole moment equal to 10-7 cm in the negative z-direction?

• Solution: In a non-uniform electric field, the force on the electric dipole is given by

F = Px + Py + Pz

= 105NC-1m-1, pz = -10-7Cm, px = 0, py = 0, =0 F = 0+0-10-7 ×€105N = -10-2N The negative sign indicates that the force is directed along negative z-axis.

Let us now calculate torque. Both and are along z-axis. ∴ sin θ = 0; ∴ τ = pEsin θ€= 0.

38. (a) A conductor A with a cavity as shown in Fig. A is given a charge Q. Show that the

entire charge must appear on the outer surface of the conductor. (b) Another conductor B with charge q is inserted into the cavity keeping B insulated from A. Show that the total charge on the outside surface of A is Q + q. (Fig. B) (c) A sensitive instrument is to be shielded from the strong electrostatic field in its environment. Suggest a possible way.

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Fig. A Fig. B

• Solution: (a) Conductor A has a cavity inside it, in which there are no charges. Consider an imaginary Gaussian surface lying within the conductor enclosing the cavity. Since there is no charge inside the surface, the electric field inside the Gaussian surface is zero.

⇒ E = 0 Thus, there is no charge inside the conductor and the entire charge appears on the outer surface of the conductor. (b) When some other charge +q is inserted into the cavity, by induction -q would appear on the inner surface of A and +q on its outer surface. Using Gauss's theorem, as done above, we can still prove that the charge inside the cavity is zero. Thus, the trial charge on the outer surface of A would be Q + q. (c) To shield a sensitive instrument from a strong electrostatic force we use the property we have shown in (a) and (b). The net electric field inside a hollow conductor is zero as flux inside the conductor is zero, a perfect hollow conductor will serve as a shield. Thus by placing the sensitive instrument inside a perfect hollow conductor we can shield it from external interference.

39. A hollow charged conductor has a tiny hole cut into its surface. Show that the

electric field in the hole is (σσσσ/2εεεε0) n, where n is the unit vector in the outward normal direction and€σ€σ€σ€σ is the surface charge density near the hole.

• Solution: We consider a hollow charged conductor of surface charged density σ€€€€with a tiny hole cut into its surface. Let q be the charge on the hollow conductor. The Gaussian surface through the point P will just enclose the hollow charged conductor. Therefore according to Gauss's theorem,

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E × area of the hole =

E × 2(4πr2) =

2E = Since q = 4πr2σ, the above equation becomes

2E =

E = n.

40. Obtain the formula for the electric field due to a long thin wire of uniform linear

charge density λλλλ without using Gauss’s law. (Hint: Use Coulomb’s law directly and evaluate the necessary integral).

• Solution:

The above figure shows an infinite line of charge whose linear charge density has a constant value λ . To calculate the field at a distance r from the line, the line is divided into a large number of very small elements of length dx and the sum of the contributions of the elements are found. The magnitude of the field contributions dE due to charge element dq (= λ dx) is given by Coulomb's law:

dE = ---------(1) The vector dE in the figure has two rectangular component dEx and dEy. The x-component of the total field E is zero because every charge element on the right of O has a corresponding element to the left of O such that their field contributions in the x-directions cancel. Thus E points entirely along the y direction. From the figure it is clear that dEy = dx cos θ . Therefore, the field E due to the entire line of charge is

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E = Ey = -----------(2) Using eqn. (1) and (2)

E = -----------(3) From the figure the relation between x and θ is given as x = rtanθ -----------(4) Differentiating, we get dx = rsec2 θ dθ -----------(5) Using eqn. (4) and (5) in (3), we get

E = (when x = 0, θ = 0 and when x = ∞ , θ = π /2) Integrating the above equation, we get

E =

= .

41. It is now believed that protons and neutrons are themselves built out of more

elementary units called quarks. A proton and neutron consist of three quarks each. Two types of quarks, so called 'up quark' denoted by u of charge +2e/3 and 'down quark' denoted by d of charge of -e/3, together with electrons build up ordinary matter. Suggest a possible quark composition of a proton and neutron.

• Solution: The charge on 'up quark' = +2e/3 The charge on 'down quark' = -e/3 The charge on the proton is +e and it is made of three quarks. i.e. +2e/3 + 2e/3 - e/3 = +e Therefore, the possible quark composition of proton is uud Neutron is neutral particle, but it is made of three quarks. i.e. +2e/3 - e/3 - e/3 = 0 hence the possible composition of neutron is udd.

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Chapter 1. Electric Charges and Fields.NCRT Question and ... 12TH/CLASS 12TH PHYS… · • Solution: The forces acting on charge q at A due to charges at B and -q at C are F12 along