Electric PotentialElectric Potential Energy

Electric Potential Energy

Work done by Coulomb force when q1 moves from a to b: b

aFE

r

drds

q2 (-)

q1 (+)ra

rb

q1 (+)

The important point is that the work depends only on the initial and final positions of q1.

In other words, the work done by the electric force is independent of path taken. The electric force is a conservative force.

b

aFE

r

drds

q2 (-)

q1 (+)

ra

rab

Electric Potential Energy

+ + + + + + + + + + + + + +

- - - - - - - - - - - - - - - - - - -

+

E

A charged particle in an electric field has electric potential energy.

It “feels” a force (as given by Coulomb’s law).

F

Electric Potential Energy

It gains kinetic energy and loses potential energy if released. The Coulomb force does positive work, and mechanical energy is conserved.

Electric Potential

Dividing W by Q gives the potential energy per unit charge.

VAB, is known as the potential difference between points A and B.

The electric potential V is independent of the test charge q0.

If VAB is negative, there is a loss in potential energy in moving Q from A to B; the work is being done by the field. if it is positive, there is a gain in potential energy; an external agent performs the work

Electric Potential

+ + + + + + + + + + + + + +

- - - - - - - - - - - - - - - - - - -

+

EF

Electric Potential

VAB is the potential at B with reference to A

VB and VA are the potentials (or absolute potentials) at B and A

If we choose infinity as reference the potential at infinity is zero;the electric potential of a point charge q is

0

1 qV r .

4 r

Electric Potential

The potential at any point is the potential difference between that point and a chosen point in which the potential is zero.

Things to remember about electric potential:

Electric potential difference is the work per unit of charge that must be done to move a charge from one point to another without changing its kinetic energy.

● Sometimes it is convenient to define V to be zero at the earth (ground).

The terms “electric potential” and “potential” are used interchangeably.

The units of potential are joules/coulomb:

1 joule1 volt =

1 coulomb

Example: a 1 C point charge is located at the origin and a -4 C point charge 4 meters along the +x axis. Calculate the electric potential at a point P, 3 meters along the +y axis.

q2q1

3 m

P

4 mx

yi 1 2

Pi i 1 2

-6 -69

3

q q qV = k = k +

r r r

1×10 -4×10= 9×10 +

3 5

= - 4.2×10 V

Thanks to Dr. Waddill for the use of these examples.

Example: how much work is required to bring a +3 C point charge from infinity to point P?

q2q1

3 m

P

4 mx

y

q3

external 3 PW q V V 0

6 3externalW 3 10 4.2 10

3externalW 1.26 10 J

The work done by the external force was negative, so the work done by the electric field was positive. The electric field “pulled” q3 in (keep in mind q2 is 4 times as big as q1).Positive work would have to be done by an external force to remove q3 from P.

Electric Potential of a Charge Distribution

Collection of charges: iP

i0 i

q1V .

4 r

Charge distribution:

P is the point at which V is to be calculated, and ri is the distance of the ith charge from P.

Pr

dq

0

1 dqV .

4 r

Potential at point P.

Electric Potential of a Charge Distribution

Example: A rod of length L located along the x-axis has a total charge Q uniformly distributed along the rod. Find the electric potential at a point P along the y-axis a distance d from the origin.

Thanks to Dr. Waddill for this fine example.

y

x

P

d

L

dq

xdx

r

=Q/L

dq=dx

2 2

dq dxdV k k

r x d

L

0V dV

y

x

P

d

L

dq

xdx

r

L L

2 2 2 20 0

dx Q dxV k k

Lx d x d

A good set of math tables will have the integral:

2 2

2 2

dxln x x d

x d

2 2kQ L L dV ln

L d

Example: Find the electric potential due to a uniformly charged ring of radius R and total charge Q at a point P on the axis of the ring.

P

R

dQ

r

xx

Every dQ of charge on the ring is the same distance from the point P.

2 2

dq dqdV k k

r x R

2 2ring ring

dqV dV k

x R

P

R

dQ

r

xx

2 2 ring

kV dq

x R

2 2

kQV

x R

Example: A disc of radius R has a uniform charge per unit area and total charge Q. Calculate V at a point P along the central axis of the disc at a distance x from its center.

P

r

dQ

xxR

The disc is made of concentric rings. The area of a ring at a radius r is 2rdr, and the charge on each ring is (2rdr).

We *can use the equation for the potential due to a ring, replace R by r, and integrate from r=0 to r=R.

ring 2 2

k 2 rdrdV

x r

P

r

dQ

xxR

R

2 2 2 2ring ring 00 0

1 2 rdr rdrV dV

4 2x r x r

R

2 2 2 2 2 22

0 0 00

QV x r x R x x R x

2 2 2 R

2

Q

R

2 22

0

QV x R x

2 R

P

r

dQ

xxR

Could you use this expression for V to calculate E? Would you get the same result as I got in Lecture 3?

MAXWELL'S EQUATION

• The line integral of E along a closed path is zero

• This implies that no net work is done in moving a charge along a closed path in an electrostatic field

MAXWELL'S EQUATION

• Applying Stokes's theorem

• Thus an electrostatic field is a conservative field

Electric Potential vs. Electric Field

• Since we have

As a result; the electric field intensity is the gradient of V

• The negative sign shows that the direction of E is opposite to the direction in which V increases

• If the potential field V is known, the E can be found

Electric Potential vs. Electric Field

Example: In a region of space, the electric potential is V(x,y,z) = Axy2 + Bx2 + Cx, where A = 50 V/m3, B = 100 V/m2, and C = -400 V/m are constants. Find the electric field at the origin

2x

(0,0,0)(0,0,0)

VE (0,0,0) Ay 2Bx C C

x

y (0,0,0)(0,0,0)

VE (0,0,0) (2Axy) 0

y

z(0,0,0)

VE (0,0,0) 0

z

V ˆE(0,0,0) 400 im