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Physics 121: Electricity & Magnetism – Lecture 5�Electric Potential

Dale E. Gary

Wenda Cao

NJIT Physics Department

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Work Done by a Constant Force

The right figure shows four situations in which a force is applied to an object. In all four cases, the force has the same magnitude, and the displacement of the object is to the right and of the same magnitude. Rank the situations in order of the work done by the force on the object, from most positive to most negative.

I, IV, III, II
II, I, IV, III
III, II, IV, I
I, IV, II, III
III, IV, I, II

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I

II

III

IV

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Work Done by a Constant Force

The work W done a system by an agent exerting a constant force on the system is the product of the magnitude F of the force, the magnitude Δr of the displacement of the point of application of the force, and cosθ, where θ is the angle between the force and displacement vectors:

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II

III

I

IV

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Potential Energy, Work and Conservative Force

Start

Then

So

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The work done by a conservative force on a particle moving between any two points is independent of the path taken by the particle.

The work done by a conservative force on a particle moving through any closed path is zero.

y_f

y_i

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Electric Potential Energy

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The potential energy of the system

The work done by the electrostatic force is path independent.
Work done by a electric force or “field”

Work done by an Applied force

U_i

U_f

U_i

U_f

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Work: positive or negative?

In the right figure, we move the proton from point i to point f in a uniform electric field directed as shown. Which statement of the following is true?

A. Electric field does positive work on the proton; And

Electric potential energy of the proton increases.

B. Electric field does negative work on the proton; And

Electric potential energy of the proton decreases.

C. Our force does positive work on the proton; And

Electric potential energy of the proton increases.

D. Electric field does negative work on the proton; And

Electric potential energy of the proton decreases.

E. It changes in a way that cannot be determined.

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E

i

f

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Electric Potential

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The electric potential energy

Start
Then
So

The electric potential

Potential difference depends only on the source charge distribution (Consider points i and f without the presence of the test charge;

The difference in potential energy exists only if a test charge is moved between the points.

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Electric Potential

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Just as with potential energy, only differences in electric potential are meaningful.

Relative reference: choose arbitrary zero reference level for ΔU or ΔV.
Absolute reference: start with all charge infinitely far away and set U_i= 0, then we have and at any point in an electric field, where W_∞ is the work done by the electric field on a charged particle as that particle moves in from infinity to point f.

SI Unit of electric potential: Volt (V)

1 volt = 1 joule per coulomb

1 J = 1 VC and 1 J = 1 N m

Electric field: 1 N/C = (1 N/C)(1 VC/J)(1 J/Nm) = 1 V/m
Electric energy: 1 eV = e(1 V)

= (1.60×10^-19 C)(1 J/C) = 1.60×10^-19J

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Potential Difference �in a Uniform Electric Field

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uphill for − q

Electric field lines always point in the direction of decreasing electric potential.
A system consisting of a positive charge and an electric field loses electric potential energy when the charge moves in the direction of the field (downhill).
A system consisting of a negative charge and an electric field gains electric potential energy when the charge moves in the direction of the field (uphill).
Potential difference does not depend on the path connecting them

downhill for + q

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Equipotential Surface

The name equipotential surface is given to any surface consisting of a continuous distribution of points having the same electric potential.

Equipotential surfaces are always perpendicular to electric field lines.

No work is done by the electric field on a charged particle while moving the particle along an equipotential surface.

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The equipotential surface is like the “height” lines on a topographic map.

Following such a line means that you remain at the same height, neither going up nor going down—again, no work is done.

Analogy to Gravity

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Work: positive or negative?

3. The right figure shows a family of equipotential surfaces associated with the electric field due to some distribution of charges. V1=100 V, V2=80 V, V3=60 V, V4=40 V. W_I, W_II, W_III and W_IV are the works done by the electric field on a charged particle q as the particle moves from one end to the other. Which statement of the following is not true?

A. W_I = W_II

B. W_III is not equal to zero

C. W_II equals to zero

D. W_III = W_IV

E. W_IVis positive

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Potential Due to a Point Charge

Start with (set V_f=0 at ∞ and V_i=V at R)

We have

Then

So

A positively charged particle produces a positive electric potential.
A negatively charged particle produces a negative electric potential

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Potential due to �a group of point charges

Use superposition

For point charges

The sum is an algebraic sum, not a vector sum.
E may be zero where V does not equal to zero.
V may be zero where E does not equal to zero.

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q

-q

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Electric Field and Electric Potential

4. Which of the following figures have V=0 and E=0 at red point?

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A

q

-q

B

q

C

D

q

E

-q

q

-q

q

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Potential due to a Continuous �Charge Distribution

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Find an expression for dq:

dq = λdl for a line distribution
dq = σdA for a surface distribution
dq = ρdV for a volume distribution

Represent field contributions at P due to point charges dq located in the distribution.

Integrate the contributions over the whole distribution, varying the displacement as needed,

So far we have discussed the electric field produced by one or a group of point charges. Now let’s consider how to find the electric field due to charge distributions. Usually this charge distribution consist of many many closely spaced point charges (perhaps millions or billions) and they are continuously distributed along some line, over some surface, or throughout some volume. For this case, we should follow this procedure to find the electric field: First, divide the charge distribution into small elements, each of them contains a small charge delta q. Next, use this equation to calculate the electric field due to one of these elements at a point P. Finally, summing the contribution of all the charge elements to find the total electric field at P. We can use this integral form to perform the calculation.

In detail, first find an expression for dq, then represent field contributions at P due to point charges dq, Take advantage of any symmetry to simplify the calculation. Finally, integrate the contributions over the whole distribution to find the electric field.

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Example: Potential Due to �a Charged Rod

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A rod of length L located along the x axis has a uniform linear charge density λ. Find the electric potential at a point P located on the y axis a distance d from the origin.

Start with

then,

So

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Potential Due to �a Charged Isolated Conductor

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According to Gauss’ law, the charge resides on the conductor’s outer surface.
Furthermore, the electric field just outside the conductor is perpendicular to the surface and field inside is zero.
Since

Every point on the surface of a charged conductor in equilibrium is at the same electric potential.
Furthermore, the electric potential is constant everywhere inside the conductor and equal to its value to its value at the surface.

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Calculating the Field from the Potential

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Suppose that a positive test charge q₀ moves through a displacement ds from on equipotential surface to the adjacent surface.
The work done by the electric field on the test charge is W = −dU = -q₀dV.
The work done by the electric field may also be written as
Then, we have

So, the component of E in any direction is the negative

of the rate at which the electric potential changes with

distance in that direction.

If we know V(x, y, z),

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Electric Potential Energy �of a System of Point Charges

Start with (set U_i=0 at ∞ and U_f=U at r)

We have

If the system consists of more than two charged particles, calculate U for each pair of charges and sum the terms algebraically.

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q₁

q₂

q₃

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Summary

Electric Potential Energy: a point charge moves from i to f in an electric field, the change in electric potential energy is
Electric Potential Difference between two points i and f in an electric field:
Equipotential surface: the points on it all have the same electric potential. No work is done while moving charge on it. The electric field is always directed perpendicularly to corresponding equipotential surfaces.
Finding V from E:
Potential due to point charges:
Potential due to a collection of point charges:
Potential due to a continuous charge distribution:
Potential of a charged conductor is constant everywhere inside the conductor and equal to its value to its value at the surface.
Calculatiing E from V:
Electric potential energy of system of point charges:

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