Solution to the lab paper PH-103MCQs Answer key1. b2. b3. e4. a5. c

6. e7. a8. a9. a10. a

Subjective:Q1.

When

And as, the potential energyQ 2. Uniform charge distributions

Charge density Using Gausss Law, we find E both inside and outside of the sphere.

For r < R, Ein is

and

(A)For r > R, Eout is

and

(B)Electric field is directed away from the center of the sphere that produces it.

(a) x = 0This point is inside sphere-1 and outside the sphere-2For E1; r = 0, Using Eq. A

For E2; r=2R, Using Eq. B

(negative x-direction)

(b) x = R/2This point is inside sphere-1 and outside the sphere-2For E1; r = R/2, Using Eq. A

(positive x-direction)For E2; r=R+R/2=3R/2, Using Eq. B

(negative x-direction)

(c) x = RThis point is on the surface of sphere-1 and sphere-2For E1; r = R, Using Eq. B

(positive x-direction)For E2; r=R, Using Eq. B

(negative x-direction)

(d) x = 3RThis point is outside the sphere-1 and lie on the surface of sphere-2For E1; r = 3R, Using Eq. B

(positive x-direction)

For E2; r=R, Using Eq. B

(positive x-direction)

Q 3. 1= 4.0 C/m2; R1=0.5 cm=0.510-2 m2= -2.0 C/m2; R2=2.0 cm =210-2 mL=6.0 cm = 610-2 m

Let P be the point on the x-axis at a distance x from the origin where electric field is zero.Electric field outside the spherical shell is

For shell-1, r=x

For shell-1, r=x-L

For the given condition at point P,E1=E2

Q 4.(i) E = 2.0104 N/Cqe = -1.6010-19 Ca = ?me = 9.110-31kg

and

Negative sign shows the direction is opposite to the direction of electric field.

(ii) (a) Co = 50 pF = 5010-12 FA = 0.35 m2d = ?

(b) = 5.6

Q 5.(a) Using Gausss Law, we find E both inside and outside of the sphere.

For spherical charge distribution

(i) r < a

(ii) a < r < b

(iii) b < r < c

(Charge inside the conductor in electrostatic equilibrium)

(iv) r > c

(b) Let Q1 be the induced charge on the inner surface of the hollow sphere. Since E = 0 inside the conductor, the total charge enclosed by a spherical surface of radius b r c must be zero.Therefore,

Let Q2 be the induced charge on the outside surface of the hollow sphere. Since the hollowsphere is uncharged, we require