Electric Potential, Electric Potential Energy,

and the Generalized Work Energy

Theorem

When an object is lifted at a constant velocity the work done equals the change in the object’s gravitational potential energy. Because the gravitational force is conservative the work done only depends on the initial and final heights and not on the path followed between the two points.

Gravitational Potential Energy and the Gravitational Potential

W =ΔGPE

W =m ghf −m g hi

W =m( ghf − g hi )

Define :

Gravitational Potential, φG

φG = g h

Then,

GPE =mφG

φG =GPEm

and

ΔGPE =mΔφG

GPE and φG .are scalarsUnits φG = Jkg

Moving an Object in a Gravitational Field

i

Reference Level

hi

f

1

3

2hf

W1 =W2 =W3 =mg hf −m ghi

Electric Potential Energy and the Electric Potential

Since the electric force (field) is conservative we can define an Electric Potential Energy, EPE and an Electric Potential, E.

The electric potential energy of a charge q located at some point in an electric field E produced by some charge distribution is defined as: EPE =qφEwhere,

φE is the electric potential at the position

.of the charge qThen,

φE =EPEq

and

ΔEPE =qΔφE

Units φE = JC =volt, V

EPE and φE .are scalars

Total Mechanical Energy and the Generalized Work-Energy Theorem

The total mechanical energy of a charged object is:

E =KE +GPE +EPE

The Generalized Work-Energy Theorem becomes:

Wnc =ΔKE + ΔGE + ΔEPE

where,

Wnc is the work done by all

nonconservative forces.

Wnc =12 mΔ(v2) +mΔφG +qΔφE

Electric Potential due to Various Charge Distributions

A) Single point charge, q1φE =

kq1r1

where r1 = distance from q1 to the

.point where the potential is being measured

NOTES :

1. Reference level is chosen to be at ∞,φE→ ∞ =0

2. ,Sign of charge q1 .must be included

B) Multiple point charges, qiφE =

kqirii,scalar

∑

where ri = distance from qi to the

.point where the potential is being measured

NOTES :1. Reference level is chosen to be at∞,φE→ ∞ =0

2. ,Sign of charges qi .must be included

C) Electric potential inside a parallel-plate capacitor carrying equal and opposite charges +/- q1

φE =q1 ⋅dεoA

where,

d = distance from the negative plate

A = surface area of plates

εo =8.85×10-12 C2

N⋅m2

NOTES :

1. Reference level is chosen to be at

the negative plate,φE→ −=0

Examples

Three point charges, q1 = 25C, q2 = -40C, and q3 = 60C lie on a straight line as shown below.What is the electric potential, E at point P, 5.5m from the origin?

0 1m 2m 3m 4m 5m 6m 7mq1 q2 q3

+ - +P

φE,P = φE,ii

∑

φE,P =kq1r1,P

+kq2r2,P

+kq3r3,P

φE,P = k(q1r1,P

+q2r2,P

+q3r3,P

)

φE,P = 9 × 109 N ⋅m2

C2 ( 25 × 10−6C5.5m

+ −40 × 10−6C1.5m + 60 × 10−6C

1.5m )

φE,P = 9 × 109 N ⋅m2

C2 (4.55 × 10−6 Cm

− 2.67 × 10−5 Cm + 4.0 × 10−5 C

m )

φE,P = 9 × 109 N ⋅m2

C2 (1.785 × 10−5 Cm )

φE,P = 1.61 × 105 N ⋅mC = 1.61×105 J

C = 1.61×105V

How much work would be required to bring a proton from infinity and place it at point P? (Ignore any changes in the proton’s KE or GPE)

Wnc =12 mΔ(v2) +mΔφG +qΔφE

0 0

W =qpΔφE

W =qp(φE,f −φE ,i )

W =qp(φE,P −φE ,∞)

W =(1.6×10−19C)(1.61×105 JC −0)

W =2.58 ×10−14 JWork must be done ON the proton to bring it from infinity and place it at point P

0 1m 2m 3m 4m 5m 6mq1 q2

+ -

Two point charges, q1 = 25C and q2 = -40C lie on a straight line as shown below.

For the electric potential to be zero the potentials due to q1 and q2 must cancel.To cancel the potentials must be equal in magnitude and opposite in sign.

There are three possibilities for the location of a point where the electric potential is zero.

1. To the right of q2, x > 4m.

2. Between q1 and q2, 0 < x < 4m.

3. To the left of q1, x < 0.

Is there a point where the electric potential is zero?

φE,1L +

φE,2L −

⎫ ⎬ ⎪

⎭ ⎪q2 > q1

r1 > r2

⎫ ⎬ ⎭

⎧ ⎨ ⎩

φE,1 < φE,2

φE,1L +

φE,2L −

⎫ ⎬ ⎪

⎭ ⎪q2 > q1

r1 < r2

⎫ ⎬ ⎭

⎧ ⎨ ⎩

φE,1 = φE,2

φE,1L +

φE,2L −

⎫ ⎬ ⎪

⎭ ⎪q2 > q1

r1 < r2

⎫ ⎬ ⎭

⎧ ⎨ ⎩

φE,1 = φE,2

0 1m 2m 3m 4mq1 q2

+ -

Between q1 and q2, 0 < x < 4m.

Since q1<q2 the point must be closer to q1.

Px 4-x

φE,P,1 = φE,P,2

k q1r1,P

=k q2r2,P

q1x =

q24 −x

q1 (4 −x) = q2 x

4 q1 − q1 x= q2 x

4 q1 = q2 x+ q1 x

x( q2 + q1 ) =4 q1

x =4 q1

q2 + q1

x = 4(25×10−6C)40 ×10−6C + 25×10−6C

x =10065 m

x =1.54m

To the left of q1, x < 0.

Any point to the left of q1 must be closer to q1.

0 1m 2m 3m 4mq1 q2

+ -Q

x

x+4

φE,Q,1 = φE,Q,2

k q1r1,Q

=k q2r2,Q

q1x =

q2x+4

q1 (x + 4) = q2 x

q1 x + 4 q1 = q2 x

4 q1 = q2 x− q1 x

x( q2 − q1 ) =4 q1

x =4 q1

q2 − q1

x = 4(25×10−6C)40 ×10−6C−25×10−6C

x =10015 m x =6.67m

0 1m 2m 3m 4mq1 q2

+ -PQ

6.67m 1.54m

5m

A =.01m2

15 ms

−4×10−7C

+ 4×10−7C

2 ×10−7C0.05kg

An object with a mass of 0.05kg and charge of 2.0x10-7C is ejected from the negative plate of a capacitor with an initial velocity of 15m/s as shown below.The plates of the capacitor are 5m apart and each has an area of .01m2. The plates carry equal and opposite charges of 4x10-7C.

Will the object reach the positive plate?

As the object moves the only forces acting on it are the gravitational and electrical forces both of which are conservative. Therefore the total mechanical energy of the object will be conserved.

The negative plate is chosen as the reference level for both the gravitational and electric potential energy. Therefore:

φG,i = 0

φE,i = 0

Ei =Ef12 mv i

2 +mφG ,i +qφE ,i = 12 mvf

2 +mφG ,f +qφE ,f

12 mv i

2 =12 mvf

2 +mghf +q′ q hf

εoA

where:q= charge on the object

′ q = charge on the capacitor plates

Let’s find the height, hf , where the object stops moving upward, vf = 0.

12 mv i

2 =mghf +q′ q hfεoA

12 mv i

2 =(mg+q ′ qεoA

)hf

hf =12 mvi

2

(mg+q ′ qεoA

)

(0.05kg)(9.8 ms2 )+ (2×10−7C)(4 ×10−7C)

(8.85 ×10−12 C2

N⋅m2 )(.01m2 )

hf =.5(0.05kg)(15 m

s )2

hf =4.035m

The object does not reach the upper plate.

What minimum initial velocity would allow the object to reach the upper plate?

12 mv i

2 =12 mvf

2 +mghf +q′ q hf

εoA

12 mv i

2 =mgd+q ′ q dεoA

vi =2(mgd+q ′ q d

εoA)

m

hf =d=5m, vf =0

vi =2{(0.05)(9.8)(5) +(2 ×10−7 )× (4×10−7 )(5)

(8.85 ×10−12 )(.01)}

(0.05kg)

vi =16.7ms