lecture 6

Work and E

nergy in Ele

ctrostatics

x

y

z

~E(~r)a

b

O

~r

OutlineI. Work done in moving a charge in an electric fieldII. Electrostatic energy of assemblies of discrete and continuous charge distributions III. Electrostatic energy of a charged solid sphere

Work and Energy in Electrostatics

Learning ObjectivesI. To learn about work done in moving a charge in an electric field.II. To learn to calculate the work done in assembling a discrete or continuous charge distribution and the corresponding electrostatic energy.


Learning OutcomesI. To be able to calculate work done in moving a charge in a given electric field.II. To be able to calculate the electrostatic energy of a charged configuration- discrete or continuous.


Work and Energy in ElectrostaticsConsider moving a test charge q from point in thepresence of an electric field . The mechanical work done on the charge q is

x

y

z

~E(~r)a

b

O

~r

So that, the work done on q is equal to

~E

W =

Z ba

~Fm ech(~r) ~dl

W = qZ ba

~E(~r) ~dl

) W = q[Vb Va] = q4Vab

a! b

~Fm ech(~r) = q ~E(~r)where

Since, 4Vab = Vb Va = Z ba

~E(~r) ~dl

Taking and , the work done on q becomes

~ra =1 V (~ra) = V (1) = 0Work and Energy in Electrostatics

where we have taken ~rb = r:

which is also equal to the potential energy of the charge q. SI units of work, potential energy and potential difference

Thus, the work done in moving a charge q from to is 1 rW = qV (~r)

Work: Joules (J)Potential energy: Coulomb-Volts=Joules Potential difference: Volts=Joules/Coulomb

W = qV (~r);

= q2V1(~r2)

W2 = q21

40

q1r12

Work and Energy in ElectrostaticsElectrostatic energy of assembly of a point charge distribution:Consider moving charges from to , one by one, as shown in the figure. What is the work done in bringing in each charge from ?

1 ~ri

1

W3 = q31

40

q1r13

+q2r23

= q3V1(~r3) + q3V2(~r3)

W1 = 0

~r12 = ~r2 ~r1; r12 = j~r12j

Work and Energy in ElectrostaticsConsider assembling three charges as shown in the figure.

x

y

z

O~r1

~r2

~r3

The position and the relative vectors are related by

The work done in assembling two charges,

V1(~r2) =1

40

q1r12

where W2 = q2

1

40

q1r12

= q2V1(~r2) ;

Similarly, for adding the third charge,W3 = q3

1

40

q1r13

+q2r23

= q3V1(~r3) + q3V2(~r3)

The total work done in assembling three charges,Work and Energy in Electrostatics

For assembling N charges, the total work done can be written as

WT O T =1

2

NXi= 1

qi

NXj = 1j 6= i

1

40

qjrij

=

1

2

NXi= 1

qiV (~ri)

Note that the second sum represents the potential at due to all the charges.

WT O T = W1 +W2 +W3 =1

40

q1q2r12

+q1q3r13

+q2q3r23

To avoid writing j>i, we can count each pair twice and then divide by 2, so that the total work done becomes

WT O T =1

40

NXi= 1

NXj = 1j > i

qiqjrij

~ri


We get for the total work done in assembling a continuous volume charge distribution,

WT O T =1

2

ZV(~r)V (~r)d

Instead of point charges, if we have a continuous charge distribution, then using

NXi= 1

qiV (~ri) =)ZVdqV (~r) =

ZV(~r)V (~r)d

It is known as the electrostatic energy of a continuous charge distribution.

ZV

~r f ~A

d =

ZVf~r ~A

d +

ZVA

~rfd =

ISf ~A ~dS


ZVf~r ~A

d =

ZV

~A ~rfd +

ISf ~A ~dSor,

Electrostatic energy of a continuous charge distribution:

WT O T =1

2

ZV(~r)V (~r)d

The above expression for electrostatic energy can be reformulated in terms of the electric field in all space as follows. Use Gauss' law to write the energy as

WT O T =02

ZV

~r ~E

V (~r)d

Use the divergence theorem to transfer the derivative from to V using and integrating over a volume V enclosed by a surface S,

~E ~r (f ~A) = f(~r ~A) + ~A (~rf)

Work and Energy in ElectrostaticsWe have,

If we let then since r!1;

The total electrostatic energy due to a localized charged distribution producing the electric field is given by ~E

Calculating Electrostatic Energy of Charged Distributions

We have the following four equations for calculating the electrostatic energy of a given charge distribution

Based on efficiency and convenience, one can use any one of these equations to calculate the electrostatic energy.

(~r):

1. WT O T = 12

ZV(~r)V (~r)d

3.

4. WT O T =Z

dqV (~r)

2.

Electrostatic Energy of a Charged Sphere

Consider a uniform volume charge density in the form of a sphere of radius R as shown in the figure. Calculate the work done in assembling the charged sphere.

(~r)

For using Eq. (1), W = 12

ZV(~r)V (~r)d

V (~r) =1

40

Q

2R

3 r

2

R2

(~r) and V (~r)

to calculate the work done in assembling the charge Q, we have to know inside the sphere.

We have, = 3Q4R3

V (~r)

R

(~r)

Substituting these values in Eq. (1),

we get,

The work done in assembling the sphere is

W =1

2

1

40

Q

2R

Z R0

3 r

2

R2

4r2dr

=1

21

0

Q

2R

R3 R

3

5

=

Q

0

R2

5=

Q2

0(4=3)R3R2

5

W =1

40

3Q2

5R

V (~r)

R

(~r)


For using Eq. (2),

All space can be divided into two regions with the electric fields in the two regions given by

W =02

ZVallspace

E2d


~E(~r)to calculate the work done, we have to know the electric field in all space.

V (~r)

R

(~r)

Substituting for in Eq. (2), ~E(~r)

R


we get,

The work done in assembling the sphere is

For using Eq. (3),


to calculate the work done, we have to know the electric field inside the volume V enclosed by the surface S.

~E(~r) and V (~r)

R

S

Substituting for in Eq. (2), we get ~E(~r) and V (~r)

or,


We see that as the work done in assembling the sphere becomes (as expected)

a!1;

V (~r)

R

(~r)

dr


W =

ZdqV (~r)

For using Eq. (4),

to calculate the work done, we consider a sphere of radius r, and then bring in a layer of charge dq from infinity to r. The charge dq is distributed uniformly over the sphere of radius r such that

dq = 4r2dr

The work done in bringing dq from infinity to r is

Since V(r) is equal to V (r) = 140

4r3

3r

dW = dqV (~r) = 4r2drr2

30=

42

30r4dr

Then the total work done is obtained by integrating from 0 to R.


Substituting for in terms of Q, we get,

the work done in assembling the sphere.

Explain the infinite electrostatic energy of a point charge q.The energy associated with the point charge is

The energy associated with the point charge is infinite.

x

y

z

O

Electrostatic Energy of a Point Charge

The problem is resolved by realizing that classical physics breaks down at length scales below the reduced Compton wavelength of the electron

Electrostatic Potential Energy and Superposition Principle

We have seen that the electric forces, fields and potentialsfollow linear superposition principle. Does electrostatic potential energy follow linear superposition principle?

~ET O T (~r) = ~E1(~r) + ~E2(~r)

Consider an electric field obtained by linearly superposing two fields,

The total electrostatic potential energy is

WT O T =02

ZVallspace

~ET O T ~ET O T d

=02

ZVallspace

f ~E1(~r) + ~E2(~r)g f ~E1(~r) + ~E2(~r)gd

6= W1 +W2 =) No linear superposition principle

Electrostatic Potential Energy and Superposition Principle

or,

Electrostatic potential energy doesn't follow linearsuperposition principle!

or,

SummaryI. We learned about the work done in moving a charge in an electric field (or potential) II. For N charges, we showed that the work done is

III. The electrostatic energy of a charge distribution can be calculated as


1.

4.3.

2.

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lecture 6

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