Electric Fields and Gauss’ Law_2014_v1

8/12/2019 Electric Fields and Gauss Law_2014_v1

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Electric Fields and Gauss Law


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#1: Electric Fields

When trying to find the force on a charge, two

or more charges are needed.

In other words, a single charge by itself, does

not produce a force on itself.

Electric fields, as we will see, allows us to

describe an electrical property for single

charges or objects.


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#2: Electric Field for a Positive Point

Charge To understand how we define electric fields (for a positive point charge to begin),

we will see how a small positive charge behaves (via the electric force) in thevicinity of a positive point charge.

We call this small positive charge a test charge, q0. This test charge is alwayspositive (when defining electric fields) and it needs to be small such that it doesnot affect the positive charge much.

If we place this test charge, q0, in the vicinity of another positive point charge, q, itwill obviously be repelled by the positive charge along the line separating the two.So depending on where the test charge is placed, the direction of the force actingon it will change, but in general, it will point away from the positive charge.

The magnitude of the force acting on the test charge, through Coulombs law is:

F = (k |q q0|)/r2

The other point to note, is that if the test charge is placed further away from the

charge (r increases), then the magnitude of the force decreases. As the test charge is moved around the point charge, this forms a pattern of force

vectors.

If we then divide out the contribution from the test charge, q0, this defines theelectric field due to the positive point charge alone.


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Charge

Force on a positive test charge in the vicinity of a positive point

charge.


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Charge

If one picks a random point around a positivepoint charge, the magnitude of the electric fieldproduced at that point due to the positive point

charge is: E = F/q0= ((k |q q0|)/ r

2)/q0 = (k |q|)/r2

The electric field is a vector at that point as welland the direction of the field is away from the

line connecting the charge and the point (samedirection as the electric force on a positive testcharge at that point)


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#5: Electric Field for a Negative Point

Charge

We can also place the test charge in the vicinity ofa negative point charge as well.

Remembering that the test charge is alwayspositive regardless of what charge it is testing,the test charge will be attracted towards thenegative charge.

The direction will change, but in general the forcevectors will point towards the negative charge.

Also, the amount or magnitude, of the electricforce on the test charges will depend on distance

just like is the case for the positive point charge.


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Charge

Force on a positive test charge in the vicinity of a negative point

charge.


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Charge

If one picks a random point around a negativepoint charge, the magnitude of the electric fieldproduced at that point due to the positive point

charge is: E = (k |q|)/r2

The electric field is a vector at that point as welland the direction of the field is towards the

charge along the line connecting the charge andthe point (same direction as the electric force on apositive test charge at that point)


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#8: Summary of Electric Field for a

Point Charges in General For both positive and negative point charges, the magnitude of the

electric field at a point some distance of r from the charge to be:

E = (k |q|)/r2

The units of electric field are N/C.

The equation for magnitude of the electric field for both positiveand negative point charges is the same, but the direction is not.

For positive charges, the field points away from the charge.

For negative charges, the field points toward the charge.

The pattern of electric field vectors are known as electric field lines.

Note that when sketching of lines, the density of lines isproportional to the magnitude of the field.


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#9: Electric Field Lines for Positive and

Negative Point Charges

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#10: Calculating Electric Field Vectors

Around Point Charges In practice, to calculate the electric field vector around a point

charge, one first measures the distance from the point charge tothe point, P, where the field is to be determined.

Then using the following equation, one can find the magnitude ofthe electric field:

E = (k |q|)/r2 Where r is the distance from the point charge to the point, P.

To determine the direction of the vector at the point, P, first draw astraight line connecting P and the charge.

For a positive point charge, the field points away from the charge at

the point P. For a negative point charge, the field points towards the charge at

the point P.

If youre not sure how to draw the direction of the electric field foreither case, just picture what direction a positive charge will moveat the point P.


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#11: Calculating Electric Field Vectors

Around Point Charges


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#12: Calculating the Total Electric Field

for Multiple Point Charges

If one is given multiple point charges and

would like to calculate the total field produced

by all the points at a point, P, then one finds

the electric field vectors produced by eachcharge at that point.

The total or net electric field vector at P, is the

vector sum of the electric fields produced byeach charge:

E= E1+ E

2+ E

3+


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#13: Calculating the Total Electric Field

for Multiple Point Charges


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#14: Relationship Between Electric

Force and Field The definition of the field comes from the force on a positive test charge, q0:

E = F/q0= ((k |q q0|)/ r2)/q0 = (k |q|)/r

2

Looking at E = F/q0we can rearrange this equation:

F = q0E

This equation would seem to apply only to a small positive test charge, however, itactually works for any charge. So, in general

F = |q| E

Thus, if at a point in space, P, we know the magnitude of the electric field (becauseits given or has been previously calculated), we can find the magnitude of theforce on the point charge using the above equation. Another, words we do nothave to use Coulombs Law if the field is known at a point in space.

It can be shown, the following is also true:

F = q E (no absolute value)

The above equation is a vector equation. It tells us a couple of important points.

If the charge is positive, then the electric force vector points in the same directionas the field. If the charge is negative, then the electric force vector points in theopposite direction of the field.

Also, if one is trying to find the components of the force from the field, then onesimply multiplies the components by the charge without the absolute value:

Fx= q Ex and Fy= q Ey


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#15: Relationship Between Electric

Force and Field


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#16: Uniform Electric Fields

There are certain objects (we will discuss them later)that can produce whats known as uniform electricfields.

A uniform electric field is a region in space where the

field is the same everywhere, both in magnitude anddirection.

The reason why uniform fields are useful to study isbecause, if a charge is placed in a uniform field, theforce on the charge is also constant because:

F = q E

If the force on a charge is constant, then by NewtonsSecond (F= m a), the acceleration is also constant:

a= F/m = (q E)/m


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#17: Uniform Electric Fields/Constant

Acceleration If the acceleration of a charge is constant, then we have seen in

Physics I, that we have equations we can use for constantacceleration to determine displacements, velocities, times, etc:

For 1 D, we have:

x= v0t + a t2

v = v0+ a t

with a = (q E)/m (could be positive or negative depending ondirection)

If the field is 2-D (similar to a projectile), then the particle behaveslike:

x= v0xt + axt2 vx= v0x+ axt

y= v0yt + ayt2

vy= v0y+ ayt

With ax= (q Ex)/m and ay= (q Ey)/m (both accelerations could be

positive or negative depending on direction)


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#18: Electric Flux

Electric flux is Latin for electric flow

Its a quantity used to measure how much electric field is

passing through a given surface.

If one is given an electric field with a magnitude of E passing

through a surface with an area of A, then the electric flux is

defined to be:

Where is the angle between the electric field and the

normal vector (a vector perpendicular to the surface)

The units for flux are (N m2)/C

Flux can be positive, negative, or zero.


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#19: Electric Flux


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#20: Rotating Area in Uniform

Field


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#21: Electric Flux Through a Closed

Surface Of course, for a single surface, there are two possible directions for the

normal vector.

One might wonder which normal vector to use when doing the flux for a

single surface. There is no correct normal vector to use for this case and

one must be told specifically which to use.

However, for physically relevant problems, the flux through a single

surface seldom occurs.

Instead, a more physically relevant problem is what is the total or net flux

through a closed a volume, also referred as a closed surface.

A enclosed volume is comprised of one or more surfaces. When finding the net flux through the volume, one first finds the

individual flux through each surface using the outward normal vector: the

normal the points away from the center of the volume.


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#22: Gauss Law

Theres a much easier method of calculating the total electric flux through

a closed surface. This is given by Gauss Law:

Gauss Law says if some arbitrarily shaped surface encloses some charges,then the net flux through the surface is equal to the total charge enclosed

by the surface, qenc, divided by a new constant, 0.

0= permittivity of free space = 1/(4 k) = 8.85 x 10-12C2/(N m2)

Most students are initially confused in using Gauss Law to calculate

electric fluxes, but its really as easy as it seems. It just amounts to findinghow much charge is enclosed by a surface to find the flux through the

enclosed surface. It does not matter the shape of the surface or where the

charges are inside the surface when determining electric flux with Gauss

Law.


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#23: Gauss Law Example: Flux

In the above picture, the 2-D lines actually represent surfaces that complete enclose the charges. Think of

strangely shaped balloons which have the charges inside of them.

For surface 1, -2Q and +Q are enclosed, so the total enclosed charge is -2Q + Q = -Q. So, qenc= -Q and the

total flux through surface 1 is E= qenc/0 = -Q/ 0 (A negative net flux, means more field is going intosurface 1 than leaving surface 1.)

For surface 2, -Q and +Q are enclosed, so the total enclosed charge is -Q + Q = 0. So, qenc= 0 and the total

flux through surface 1 is E= qenc/0= 0 (As much field is entering surface 2 as is leaving surface 2.)

For surface 3, -Q, +Q, and -2Q are enclosed, so the total enclosed charge is -Q + Q2Q = -2 Q. So, qenc= -

2Q and the total flux through surface 1 is E= qenc/0= -2Q/ 0 (Twice the flux as surface 1.)

For surface 4, no charge is enclosed, qenc= 0 and the total flux through surface 1 is E= qenc/0= 0 (Once

again, its the same concept as surface 2.)

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#24: Electric Fields for Charge

Distributions

Up until this point, weve only been talking about point charges which are

tiny little charges.

However, in nature, objects which are not point charges can also acquire

charge. For instance, charge can be placed on a large sheet of metal or ona metal sphere. These objects are called charge distributions because

charge is not located in just one point, instead its distributed over these

objects.

Calculus can be used to determine the electric fields of charge

distribution, but an even easier method of determining electric fields is touse Gauss Law again.

Thus, Gauss Law can be used to determine electric flux (which weve

already seen) and electric fields for symmetrical charge distributions.

However, the methods used for each are quite different.

#25 U i G L t C l l t


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#25: Using Gauss Law to Calculate

Electric Fields for Symmetric

Charge Distributions Gauss law was initially developed to calculate a flux and that is all it can tell us directly, so to determine

electric fields, we have to be creative.

First of all, we must measure some sort of flux, or else we can not use Gauss law at all.

The flux that will be measured is the flux through what is known as a Gaussian surface.

A Gaussian surface is a surface thats drawn concentric to the charge distribution in question and it is alsoa surface which mimics the geometry of the charge distribution in question. Another words, when

trying to determine the electric field of a metal sphere, one would draw a spherically shaped Gaussian. If

one were trying to measure the electric field of something cylindrical, then one would draw a cylindrically

shaped Gaussian.

It turns out, through Calculus, that if the shape of the Gaussian is the shape of the charge distribution than

at every point on the surface the outward normal either points with or against the field produced by the

charge distribution, so essentially the angle between them is always zero. If this is the case, then through

original definition of flux (slide #18).

E= E A, where A is the surface area of the Gaussian.

If the Gaussian is centered, then the surface is equidistance to the charge distribution, so the E in the

equation above is simply constant at the Gaussian surface.

Thus, one can then use Gauss Law as well and find that E= qenc/0, where now qencis the charge

enclosed by the Gaussian. So, then we have:

E A = qenc/0 and thus, the magnitude of the electric field, is: E = qenc/(0 A)


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#26: Using Gauss Law to Calculate

the Electric Field of a Point Charge

In the above picture, since a point charge looks like a tiny sphere, the Gaussian surface is drawn as a

sphere centered on the charge.

The flux through the Gaussian (when it is centered) is:

E= E A, and A is the surface area of a spherical Gaussian surface. The surface area of a spherical surface

with a radius of r (the Gaussian has a radius of r), is:

A = 4 r2, so:

E= E 4 r2

From Gauss law, since the charge enclosed by the Gaussian is q, we also have:

E= qenc/0 = q /0

Equating the two fluxes to one another:

E 4 r2 = q /0and solving for E:

E = q/(4 r2 0) = 1/(4 0) (q/r2), but k = 1/(4 0) , so E = k q/r2as it should be.

#27: Us ng Gauss Law to Ca cu ate


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#27: Us ng Gauss Law to Ca cu atethe Electric Field of an Infinite

Sheet of Charge

Suppose we have an infinite sheet which is positively charged with charge spread out evenly all over it. This is called a

uniform charge distribution when the charge is spread out evenly. This means if we randomly choose two different, but

equal areas of the sheet, the amount of charge in each area is the same.

The we draw a cylindrical (could be rectangular too) surface through part of sheet in order to find the flux through the

cylinder.

Since the charge is positive, the field just points straight away from the surface on both sides, since there is positive charge

on each side of the sheet.

Hence the flux, since the outward normal also points in the same direction as the field, through the right side (top) of the

cylinder as seen above is E A, where A is the area to the right side of the cylinder.

The flux through the left side is also E A as it is for the right side. This is because once again the outward normal and E point

in the same direction.

For the sides of the cylinder, there is no flux because no field passes through them. So the total flux is:

E= E A + E A = 2 E A

We do not know how much charge is enclosed by the Gaussian, but we can say, in general, that Q is enclosed.

By Gauss Law: E= qenc/0 so 2 E A = Q/0, now solve for E:

E = Q 2 A

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#28: Using Gauss Law to Calculate the


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#28: Using Gauss Law to Calculate the

Electric Field of an Infinite Sheet of

Charge

From the last slide, we se that:

E = Q/(2 A 0)

However, if the charge is spread out uniformly all over the sheet, then the quantity Q/A is constant. Another words if we

pick an area of 0.5 m2, there might be 2 micro-Coulombs in that area and 2 micro-Coulomb/0.5 m2= 4 micro-Coulomb/m2.

If we pick an area of 1 m2, then the charge in it would be twice as much, or 4 micro-Coulombs, but 4 micro-Coulomb/1 m2=

4 micro-Coulomb/m2.

Another words, the charge per area is constant for the sheet. We call this quantity charge per unit area and it is used to

characterize infinite sheets instead of using charge as we do for point charges. So, Charge per unit area = = Q/A

So in terms of :

E = (Q/A) (1/2 0) = ||/(2 0) (|| in case the sheet was negative.)

The above equation represents the amount or magnitude E. The direction is perpendicularly away from the sheet for

positive charges and perpendicularly towards the sheet for negative charges.

Also, notice there is no distances in the equation for E. For infinite sheets, the field around them is constant. It does notdepend on distance on from the sheet.

For finite sheets, the field around the edges is not constant, but it is in the middle of the sheet. Even for finite sheets, we

will use the above equation for the electric field.


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#29: Parallel Plate Capacitors If we take two infinite sheets which are parallel and

of opposite charges and place them near each other,

they form what is known as a parallel plate

capacitor.

Keeping in mind the field each sheet produces does

not depend on distance, above both sheets, the

fields from each of the infinite sheets cancels out

since they are in opposite directions and same

magnitude. Beneath both sheets the fields also cancel out since

they are in opposite directions as well.

In between, both sheets the point in the same

direction, so the magnitude of the fields add:

E = ||/(2 0) + ||/(2 0) = ||/(0)

So the magnitude of the field is just twice the

amount of one single sheet. It is also constant as it

was for a single sheet.

The direction of the field is obviously away from the

positive sheet, towards the negative sheet.

Once again, for finite capacitors, the field is not

different near the edges, but in the middle of the

sheets, the field is equal to the equation above. So

even for finite capacitors, we will use the aboveequation for the field.


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#30: Electric Field For Parallel Plate

Capacitors

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Electric Fields and Gauss’ Law_2014_v1