Electric Potential

CONSERVATIVE FORCES

A conservative force “gives back” work that has been done against it

Gravitational and electrostatic forces are conservative

Friction is NOT a conservative force

CONSERVATIVE FORCES

A conservative force “gives back” work that has been done against it

When we lift a mass m from ground to a height h,

the potential energy of the mass increases by mgh.

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If we release the mass, it falls, picking up kinetic

energy (or speed). As the mass falls, the potential

energy is being converted into kinetic energy.

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By the time it reaches the ground, the mass has

acquired a kinetic energy ½ mv² = mgh, and it’s

potential energy is zero.

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The gravitational force ‘gave back’ the work that

we did when we lifted the mass.

CONSERVATIVE FORCES

A conservative force “gives back” work that has been done against it

The gravitational force is a conservative force.

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The electric force is a conservative force as well.

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We will be able to define a potential energy

associated with the electric force. A charge will

have potential energy when in an electric field.

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Work done on the charge (by an external agent,

or by the field) will result in changes in the

potential energy of the charge.

CONSERVATIVE FORCES

A conservative force “gives back” work that has been done against it

POTENTIAL ENERGY

POTENTIAL ENERGY

Potential energy is a relative quantity, that means, it is always the

difference between two values, or it is measured with respect to a

reference point (usually infinity).

We will always refer to, or imply, the change in potential energy

(potential energy difference) between two points.

POTENTIAL ENERGY IN A CONSTANT FIELD E

The potential energy difference between A and B

equals the negative of the work done by the field

as the charge q is moved from A to B

ΔU_AB = U_B – U_A = -W_AB = -F_E L = q E L

POTENTIAL ENERGY IN A CONSTANT FIELD E

E

Potential energy difference between A and B

ΔU_AB = U_B – U_A = - ∫ q E.dl

But E = constant, and E.dl = -1 E dl, then:

ΔU_AB = - ∫ q E.dl = ∫ q E dl = q E ∫ dl = q E L

POTENTIAL ENERGY IN A CONSTANT FIELD E

The potential energy difference between A and B

equals the negative of the work done by the field

as the charge q is moved from A to B

ΔU_AB = q E L when the +q charge is moved against the field

ELECTRIC POTENTIAL DIFFERENCE

The potential energy ΔU depends on the charge being moved.

In order to remove this dependence, we introduce the concept

of electric potential ΔV

ΔV_AB = V_B – V_A

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Electric potential difference between the points A and B

ELECTRICAL POTENTIAL DIFFERENCE

The potential energy ΔU depends on the charge being moved.

In order to remove this dependence, we introduce the concept

of electrical potential ΔV

ΔV_AB = ΔU_AB / q

Electrical Potential = Potential Energy per Unit Charge

ΔV_AB = Electrical potential difference between the points A and B

ELECTRIC POTENTIAL IN A CONSTANT FIELD E

The electric potential difference between A and B equals the negative of the work per unit charge, done by the field,

as the charge q is moved from A to B

ΔV_AB = V_B – V_A = -W_AB /q = qE L/q = E L

ELECTRICAL POTENTIAL IN A CONSTANT FIELD E

The electrical potential difference between A and B equals the work per unit charge necessary, for an external agent, to move a charge +q from A to B

ΔV_AB = V_B – V_A = -W_AB /q = - ∫ E.dl

But E = constant, and E.dl = -1 E dl, then:

ΔV_AB = - ∫ E.dl = ∫ E dl = E ∫ dl = E L

ΔV_AB = ΔU_AB / q

ΔU_AB = q E L

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ELECTRIC POTENTIAL

IN A CONSTANT FIELD E

ΔV_AB

ΔV_AB = ΔU_AB / q

POTENTIAL ENERGY

IN A CONSTANT FIELD E

ΔU_AB

UNITS

Potential Energy ΔU: [Joule] ≡ [N m]

(energy = work = force x distance)

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Electric Potential ΔV: [Joule/Coulomb] ≡ [Volt]

(potential = energy/charge)

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Electric Field E: [N/C] ≡ [V/m]

(electric field = force/charge = potential/distance)

Cases in Which the Electric Field E is not Aligned with dL

•

A

B

E

•

θ

E . dl = E dl cos θ ⇒ ΔV_AB = - E cos θ ∫ dl = - E L cos θ

Since F = q E is conservative, the field E is conservative.

Then, the electrical potential difference does not depend

on the integration path.

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One possibility is to integrate along the straight line AB.

This is convenient in this case because the field E is constant, and the angle θ between E and dL is constant.

A

B

Cases in Which the Electric Field E is not Aligned with dL

•

•

A

B

E

•

C

X

Another possibility is to choose a path that goes from A to C, and

then from C to B

ΔV_AB = ΔV_AC + ΔV_CB ΔV_AC = E X ΔV_CB = 0 (E ⊥ dL)

Thus, ΔV_AB = E X but X = L cos ϕ = - L cos θ

θ

ϕ

L

Equipotential Surfaces (lines)

Since the field E is constant

All the points along the dashed line,

at X, are at the same potential.

The dashed line is an

equipotential line

Then, at a distance X from plate A

Equipotential Surfaces (lines)

X

It takes no work to move a charge

at right angles to an electric field

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E ⊥ dL ⇒ ∫ E•dL = 0 ⇒ ΔV = 0

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If a surface (line) is perpendicular to

the electric field, all the points in

the surface (line) are at the same

potential. Such surface (line) is called

EQUIPOTENTIAL

EQUIPOTENTIAL ⊥ ELECTRIC FIELD

Potential Due to a Point Charge

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

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• We have

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• Then

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• So

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

Potential due to �a group of point charges

• Use superposition

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• For point charges

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• 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.

October 3,

Potential due to a Continuous �Charge Distribution

• Find an expression for dq:
• dq = λdl for a line distribution
• dq = σdA for a surface distribution
• dq = ρdV for a volume distribution

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

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

Example: Potential Due to �a Charged Rod

• 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.

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• Start with

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• then,

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• So

Potential Due to �a Charged Isolated Conductor

• 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.

Calculating the Field from the Potential

• 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 = -q₀ dV.
• The work done by the electric field may also be written as
• Then, we have

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• 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), we can calculate E

Thanks