electric potential chapter 25. electric potential difference the fundamental definition of the...

Electric Potential

Chapter 25

ELECTRIC POTENTIAL DIFFERENCE

The fundamental definition of the electric potential V is given in terms of the electric field:

VAB = - AB E · dl

VAB = Electric potential difference between the points A and B = VB-VA.

This is not the way we will usually calculate electric potentials, but we will explore this in a couple of simple examples to understand it better.

A

B

Constant electric field

VAB = - AB E · dl•

A

B

E

•

E · dl = E dl cos(-) VAB = -E cos(-) dl = E L cos

The electric potential difference does not depend on the integration path. So pick a simple path.

One possibility is to integrate along the straight line AB.This is easy in this case because E is constant and the angle between E and dl is constant.

L

dl

• •A

B

E

•

Cd

This line integral is the same for any path connecting the same endpoints. For example, try the two-step path A to C to B.

VAB = VAC + VCB VAC = E d VCB = 0 (E dl)

Thus, VAB = E d but d = L cos VAB = E L cos

L

VAB = - AB E · dl

Constant electric field

Notice: the electric field points downhill

Equipotential Surfaces (lines)

For a constant field E all of the points along the vertical line A are at the same potential.

EA


For a constant field E all of the points along the vertical line A are at the same potential. Pf: Vbc=-∫E·dl=0 because E dl. We can say line A is at potential VA.

EA

b

c



EA x

The same is true for any vertical line: all points along it are at the same potential. For example, all points on the dotted line a distance x from A are at the same potential Vx, where VAx = E x



EA x

A line (or surface in 3D) of constant potential is known as anEquipotential

The same is true for any vertical line: all points along it are at the same potential. For example, all points on the dotted line a distance x from A are at the same potential Vx, where VAx = E x

We can make graphical representations of the electric potential in the same way as we have created for the electric field:

Equipotential Surfaces

Lines of constant E

We can make graphical representations of the electric potential in the same way as we have created for the electric field:


Lines of constant ELines of constant V(perpendicular to E)


Equipotential plots are like contour maps of hills and valleys. The electric field is the local slope, and points downhill.

Lines of constant ELines of constant V(perpendicular to E)

It is sometimes useful to draw pictures of equipotentials rather than electric field lines:


How do the equipotential surfaces look for:(a) A point charge?

(b) An electric dipole?

+

+ -

E

Equipotential plots are like contour maps of hills and valleys. The electric field is the local slope, and points downhill.

Point Charge q

The Electric Potential

b

aq

What is the electrical potential differencebetween two points (a and b) in the electricfield produced by a point charge q?

Electric Potential of a Point Charge

Place the point charge q at the origin. The electric field points radially outwards.


c

b

aq

Choose a path a-c-b.

Vab = Vac + Vcb Vab = 0 because on this path

€

rE ⊥d

r r

€

− r

E (r r ) • d

r r

r r c

r r b

∫ = − E(r)drra

rb

∫ = −kqdr

r2ra

rb

∫ =kq

r ra

rb

Vbc =


€

Vab = kq1

rb

−1

ra

⎛

⎝ ⎜

⎞

⎠ ⎟


From this it’s natural to choose the zero of electric potential

to be when ra

Letting a be the point at infinity, and droppingthe subscript b, we get the electric potential:

When the source charge is q,and the electric potential isevaluated at the point r.

c

b

aq


€

Vab = kq1

rb

−1

ra

⎛

⎝ ⎜

⎞

⎠ ⎟

€

V (r r ) =

kq

r


This is the most important thing to know about electric potential: the potential of a point charge q.

q

Remember: this is the electric potential with respect to infinity – we chose V(∞) to be zero.


€

V (r r ) =

kq

r

Never do this derivation again. Instead, know this simple result by heart:

r

Potential Due to a Group of Charges

• The second most important thing to know about electric potential is how to calculate it given more than one charge

• For isolated point charges just add the potentials created by each charge (superposition)

• For a continuous distribution of charge …

Potential Produced by aContinuous Distribution of Charge

dq

A

dVA = k dq / rr

VA = dVA = k dq / r

Remember:

k=1/(40)

In the case of a continuous charge distribution, divide the distribution up into small pieces and then sum (integrate) the contribution from each bit:

Example: a disk of charge

Suppose the disk has radius R and a charge per unit area .Find the potential at a point P up the z axis (centered on the disk).

Divide the object into small elements of charge and find thepotential dV at P due to each bit. For a disk, a bit (differential of area) is a small ring of width dw and radius w.

dw

P

r

Rw

z

dq = 2wdw

€

dV =1

4πε0

dq

r=

1

4πε0

σ 2πwdw

w2 + z2

∴V = dV =σ

2ε0

∫ (w2

0

R

∫ + z2)− 1

2wdw

V =σ

2ε0

( R2 + z2 − z)

Field and Electric Potential

Remember from calculus that integrals are antiderivatives.

€

f (x) = g(y)dyx0

x

∫

By the fundamental theorem of calculus you can “undo” the integral:

€

f '(x) = g(x) ⇒ g(x) = f '(x)Given


Very similarly you can get E(r) from derivatives of V(r).


€

f (x) = g(y)dyx0

x

∫


€

f '(x) = g(x) ⇒ g(x) = f '(x)Given



€

V (r r ) = −

r E ⋅d

r l r

r 0

r r

∫


€

f (x) = g(y)dyx0

x

∫


€

f '(x) = g(x) ⇒ g(x) = f '(x)Given

Choose V(r0)=0. Then



is the gradient operator

€

V (r r ) = −

r E ⋅d

r l r

r 0

r r

∫


€

f (x) = g(y)dyx0

x

∫


€

f '(x) = g(x) ⇒ g(x) = f '(x)

€

rE (

r r ) = −

r ∇V (

r r )

= −∂V

∂xˆ i +∂V

∂yˆ j +∂V

∂zˆ k

⎛

⎝ ⎜

⎞

⎠ ⎟

The third most important thing to know about potentials.

Given

Choose V(r0)=0. Then

Force and Potential Energy

This is entirely analogous to the relationship between a conservative force and its potential energy.

can be inverted:

€

U(r r ) = −

r F ⋅d

r l r

r 0

r r

∫

€

rF (

r r ) = −

r ∇U(

r r )

In a very similar way the electric potential and field are related by:

can be inverted:

€

V (r r ) = −

r E ⋅d

r l r

r 0

r r

∫

€

rE (

r r ) = −

r ∇V (

r r )

The reason is that V is simply potential energy per unit charge.


• Suppose the disk has radius R and a charge per unit area .• Find the potential and electric field at a point up the z axis.• Divide the object into small elements of charge and find the• potential dV at P due to each bit. So here let a bit be a small • ring of charge width dw and radius w.

dw

P

r

Rw

z

dq = 2wdw

€

dV =1

4πε0

dq

r=

1

4πε0

σ 2πwdw

w2 + z2

∴V = dV =σ

2ε0

∫ (w2

0

R

∫ + z2)− 1

2wdw

V =σ

2ε0

( R2 + z2 − z)

This is easier than integrating over thecomponents of vectors. Here we integrateover a scalar and then take partial derivatives.


dw

P

r

Rw

z

V(z) =20

( R2 + z2 −z)

By symmetry one sees that Ex=Ey=0 at P.Find Ez from

Ez =−∂V∂z

=−20

zR2 + z2

−1 ⎛ ⎝ ⎜ ⎞

⎠

Example: point charge

Put a point charge q at the origin.

qFind V(r): here this is easy: r

V(

r r ) =k

qr




V(

r r ) =k

qr

Then find E(r) from the derivatives:

€

rE (

r r ) = −(ˆ i ∂∂x + ˆ j ∂∂y + ˆ k ∂∂z)V(x,y,z)




V(

r r ) =k

qr


r E (

r r ) =−(ˆ i∂∂x+ ˆ j∂ ∂y+ ˆ k∂ ∂z) ( ,V x ,y )z

Derivative:∂∂x

1

r=∂

∂x

1

x2 + y2 + z2= −

1

2

2x

x 2 + y2 + z2( )3 2




V(

r r ) =k

qr


r E (

r r ) =−(ˆ i∂∂x+ ˆ j∂ ∂y+ ˆ k∂ ∂z) ( ,V x ,y )z

Derivative:∂∂x

1

r=∂

∂x

1

x2 + y2 + z2= −

1

2

2x

x 2 + y2 + z2( )3 2

r E (

r r ) =kq

x i + y j + zˆ kr 3 =kq

r r

r3=kq

ˆ rr 2So:

Energy of a Charge Distribution

How much energy ( work) is required to assemble acharge distribution ?.

CASE I: Two Charges

Bringing the first charge does not require energy ( work)


How much energy ( work) is required to assemble acharge distribution ?.

CASE I: Two Charges

Bringing the first charge does not require energy ( work)

Bringing the second charge requires to perform workagainst the field of the first charge.

r

Q1 Q2


CASE I: Two Charges

Bringing the second charge requires to perform work against the field of the first charge.

r

Q1 Q2

W = Q2 V1 with V1 = (1/40) (Q1/r)

W = (1/40) (Q1 Q2 /r) = U

U = (1/40) (Q1 Q2 /r)U = potential energy of two point charges


CASE II: Several Charges

U12 = (1/40) (Q1 Q2 /r)U12 = potential energy ofa pair of point charges

a

Q Q

Q Q

How much energy is stored in this square chargedistribution?, or …What is the electrostatic potential energy of the distribution?, or …How much work is needed to assemble thischarge distribution?

To answer it is necessary to add up the potential energy ofeach pair of charges U = Uij

CASE III: ParallelPlate Capacitor


-Q

+Q

fields cancel

fields cancel

fieldsadd

dE

A

Electric Field E = / 0 = Q / 0 A ( = Q / A)

Potential Difference V = E d = Q d / 0 A



-Q

+Q

fields cancel

fields cancel

fieldsadd

dE

A

Now, suppose moving an additional very small positive charge dq from the negative to the positive plate. We need to do work. How much work?

dW = V dq = (q d / 0 A) dq

We can use this expression to calculate the total work needed tocharge the plates to Q, -Q



-Q

+Q

fields cancel

fields cancel

fieldsadd

dE

A

dW = V dq = (q d / 0 A) dq

The total work needed to charge the plates to Q, -Q, is given by:

W = dW = (q d / 0 A) dq = (d / 0 A) q dq

W = (d / 0 A) [Q2 / 2] = d Q2 / 2 0 A



-Q

+Q

fields cancel

fields cancel

fieldsadd

dE

A

W = U = d Q2 / 2 0 A

The work done in charging the plates ends up as stored potential energy of the final charge

distribution

Where is the energy stored ? The energy is stored in the electric field



-Q

+Q

fields cancel

fields cancel

fieldsadd

dE

A

U = d Q2 / 2 0 A = (1/2) 0 E2 A d

The energy U is stored in the field, in the region between the plates.

E = Q / (0 A)

The volume of this region is Vol = A d, so we can define the energy density uE as:

uE = U / A d = (1/2) 0 E2


.

uE = U / A d = (1/2) 0 E2 Electric Energy Density

Although we derived this expression for the uniform field of a parallel plate capacitor, this is a universal expression valid for any electric field.

When we have an arbitrary charge distribution, we can use uE

to calculate the stored energy U

dU = uE d(Vol) = (1/2) 0 E2 d(Vol) U = (1/2) 0 E2 d(Vol)

CASE IV: ArbitraryCharge Distribution

[The integral covers the entire region in which the field E exists]

A Shrinking Sphere

A sphere of radius R1 carries a total charge Q distributed evenlyover its surface. How much work does it take to shrink the sphereto a smaller radius R2 ?.

electric potential chapter 25. electric potential difference the fundamental definition of the...

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