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Electric Potential Energy

Electric Potential Energy

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

r

b

dr

q1(+)

ds

FE

rb

q1(+)

a

ra

q2 (-)

r

b

dr

q1 (+)

ds

FE

rab

a

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

ra

q2 (-)

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

Electric Potential Energy

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

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

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

F

E

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

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

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.

Electric Potential

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

If VABis 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

F

E

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

VABis the potential at B with reference to A

VBand VAare 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

The potential at any point is the potential differencebetween 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:

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.

y

P

3 m

x

q2

q1

4 m

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

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

0

y

q3

P

3 m

x

q2

4 m

q1

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:

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

Charge distribution:

dq

r

P

Potential at point P.

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.

y

=Q/L

P

r

dq=dx

d

dq

x

dx

x

L

Thanks to Dr. Waddill for this fine example.

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.

dQ

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

r

R

P

x

x

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.

dQ

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).

r

P

x

x

R

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

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

dQ

r

P

x

x

R

- 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 closedpath in an electrostatic field

- 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 directionof E is opposite to the direction in which V increases

Electric Potential vs. Electric Field

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

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

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