CHAPTER 23

ELECTRIC POTENTIAL

• Potential difference and electric field

• Potential difference between two parallel plates

• Potential due to a single point charge

• Potential due to a collection of charges• Work done bringing charges together

• Potential for continuous charge distributions• Charged, hollow sphere• Uniformly charged ring

• Equipotential surfaces

• Electrostatic potential energy

EFM08AN1.MOV

Work done and potential energy.

When a system, e.g., you, a compressed spring, an electric field, does (positive) work, it loses potential energy and the amount of work done by the system is the same as the loss of potential energy:

δW = −δU.

Work done and potential energy ...

Let’s look at the similarity between electric and gravitational fields. The work done by the g-field in moving the mass from a → b is:

δW =

! F • d! ℓ

a

b∫ = mgδℓ ( > 0), i.e., positive work.

The change in potential energy of the mass in the gravitational field in moving from a → b is:

δU = Ub − Ua = −mgδℓ ( < 0), i.e., a loss.

∴δW = −δU = −(Ub − Ua ).Note: the work done by the g-field ( δW) in moving the mass from a → b is the same as the work you do in raising the mass from b→ a.

! g

δ! ℓ

! F = m

! g

a

b +

+

! E

- - - -- - - - - -

- -

! F = q

! E

δ! ℓ

If a (positive) charge moves through a

displacement δ! ℓ in a field

! E , the work

done by the E-field in moving the charge from a → b is

δW =! F • δ! ℓ = −δU,

where δU (= Ub − Ua ) is the change in potential energy

of the charge in the ! E -field. But

! F = q#

! E , so the change in

potential energy of the charge in the field is:

δU = (Ub − Ua ) = −! F •δ! ℓ = −q#

! E •δ! ℓ .

We define the electric potential (V) as the potential energy per unit charge,

i.e., V =

Uq#

.

So the potential difference between b and a

δV = Vb − Va = Vba

is given by: δV =

δUq#

= −! E • δ! ℓ .

Since δV < 0, Va > Vb, i.e., the charge moves from a

higher to a lower potential.

δ! ℓ

a ! F

b

q#+

We will work problems in Cartesian (x,y,z) and polar

(r,θ,φ) coordinate systems.

[1] Cartesian coordinates:

! E ⇒ (Ex ,Ey,Ez) and δ

! ℓ ⇒ (δx,δy,δz)

δV = −! E • δ! ℓ

= −(Exˆ i + Ey

ˆ j + Ezˆ k ) • (δxˆ i + δy j + δzˆ k )

i.e., δV = −(Ex.δx + Ey.δy + Ez.δz).

∴Ex = −

∂V∂x

, Ey = −∂V∂y

, Ez = −∂V∂z

i.e.,

! E = −

∂V∂x

ˆ i +∂V∂y

ˆ j + ∂V∂z

ˆ k ⎛ ⎝ ⎜

⎞ ⎠ ⎟ .

These are the basic relationships between the electric

field, ! E , and the electric potential, V, in Cartesian

coordinates.

UNITS:If E ⇒ N/C and r ⇒ m

then: V ⇒ volts (V).

But, by definition:

Ex = −

dVdx

, etc.,

then: E ⇒ V/m, which means that N/C ≡ V/m.

(It is more usual to use V/m as the unit of electric field.)

DISCUSSION PROBLEM:

When you charge a balloon by friction, its electric potential is ~10,000V, but it is safe to handle! And yet, a typical socket operates at a potential of 120V but will give you a (potentially!) fatal shock.

* What’s the difference? * Why is the socket more “shocking”?

~10,000V 120V

Conventional definition of work done in an electric field ...

The work done by the field in moving the charge from a → b is

δW = −δU = −(Ub − Ua )

= −(q!Vb − q!Va ) = q!(Va − Vb ).

(Remember, by definition ⇒ V = U

q!)

Conventionally, when a charge moves from a → b we write the work done by the field as:

δW = q!(Va − Vb) = q!Vab,

where Vab is the potential difference between the start

point (a) and the end point (b). (Note also if the charge was released and free to move in the field, δK = δW.)

Therefore, the work done by you in moving a charge from a → b is:

δW = −q!Vab = −q!(Va − Vb).

δ" ℓ

a " F

b

q!+

Find the potential difference

between a and b (displacement ! ℓ )

in a field ! E = Ex

ˆ i produced

between two parallel, infinitely large charged plates, spaced a distance d apart. Along the displacement, the change in potential is:

dV = −! E • d! ℓ = −Ex

ˆ i • (dxˆ i + dy j ) = −Exdx.

∴Vb − Va = dV

a

b∫ = −Ex dx

a

b∫ = −Exx = −

σε#

x

∴Vb = Va −

σε#

x

Example using Cartesian coordinates ...• Potential between two parallel charges plates

i

j ! ℓ = xˆ i + y j

+− ΔV 1 2

d

! E = Ex

ˆ i

! ℓ

1 2

a

b

d

Vb = Va −

σε!

x

For the pair of plates

(V2 − V1) = ΔV = −

σε!

d,

i.e., V1 > V2.

" E = Ex

ˆ i

d

+σ −σ

V1 V2

1 2

• ΔV is independent of y, it depends only on σ and d.

Thus, ΔV is the same between any point on plate 1 and

any point on plate 2 . This means that the potential is

constant over an infinitely charged plate.

• The work done by the field in moving a charge q from a → b is:

δW = q(V1 − V2) > 0,

so a +ve charge moves from a position of higher potential ( V1) to lower potential ( V2).

V V1

V2

x d ΔV

Question 23.1: Shown here is a plot of electric potential versus distance in a region where there is an electric field. Where is the magnitude of the electric field greatest?

V

x a

b c

de

V

x a

b c

de

By definition:

Ex = −

dVdx

,

so Ex is a maximum at d and e.

Question 23.2: The facing surfaces of two large parallel conducting plates separated by 10.0 cm have uniform surface charge densities +σ and −σ. The difference in potential between the plates is 500V.

(a) Which plate has the higher potential?

(b) What is the magnitude of the electric field between the plates?

(c) An electron is released from rest next to the negatively charged plate. What is the work done by the electric field as the electron moves from the negatively charged plate to the positively charged plate?

(d) What is the change in potential energy of the electron as it moves from one plate to the other?

(e) What is its kinetic energy when it reaches the positively charged plate?

(a,b) Since the field moves a +ve charge from the +σ plate to the −σ plate, the +σ plate is at the higher potential.

ΔV = (V1 − V2 ) = 500V = −E.d,

∴E = ΔV d = 500 0.1= 5000 V/m.

(c) Work done by the field is W2→1 = qV21.

But V21 = (V2 − V1) = −500 V.

∴W2→1 = −1.6 ×10−19C × (−500 V) = 8.0 ×10−17 J.

(d) The change in potential energy of the electron:

ΔU = U1 − U2 = qV1 − qV2 = q(V1 − V2)

= −1.6 ×10−19C × 500 V = −8.0 ×10−17 J.

(e) Mechanical energy is conserved:

∴ΔK = −ΔU = 8.0 ×10−17 J.

ΔK =

12

m(v2 − v!2). ∴v =

2ΔKm

=1.33 ×107 m/s.

Note also, from the work - kinetic energy theorem:

ΔK = W2→1 = qV21.

10 cm

+σ −σ

e− 1 2

0

Question 23.3: The electric field in a certain region of 3-dimensional space varies as:

V = (4xz − 5y + 3z2) volts.

What is the electric field at (2,−1,3), where all distances are in meters.

From earlier, the components of the electric field are related to the electric potential in the (x, y, z) coordinate system by the relationships:

Ex = −

∂V∂x

, Ey = −

∂V∂y

, Ez = −

∂V∂z

.

Given V = (4xz − 5y + 3z2), we have

Ex = −

∂V∂x

= −4z,

Ey = −

∂V∂y

= 5,

Ez = −

∂V∂z

= −4x − 6z.

So, at ! r = (x, y,z)⇒ (2,−1,3) m,

Ex = −4z = −12 V/m,

Ey = 5 V/m,

Ez = −4x − 6z = −26 V/m.

∴! E = −12ˆ i + 5ˆ j − 26 ˆ k ( ) V/m.

[2] Polar coordinates

If the electric field has radial symmetry, i.e., it depends only on

! r , e.g., a point charge, then

! E (r)⇒ Erˆ r = k

Qr2 ˆ r .

For a radial displacement d! r (in

the r direction):

dV(r) = −! E (r)• d! r = −Erˆ r • d! r .

But d! r // ˆ r ,

∴ˆ r • d! r = ˆ r d! r cos0 = dr.

Therefore, potential difference between radii r2 and r1 is

V21 = V(r2) − V(r1) = − Er

r1

r2∫ dr.

But dV(r) = −Erdr, so the radial electric field is

Er = −

dV(r)dr

.

Again, we have simple relations between the electric field

Er and the electric potential V(r).

+ d! r

r

= 1

Example ... electric potential for a point charge:

For a small displacement d! r in the radial direction ( r ),

the change in potential is:

dV(r) = −

! E (r) • d! r = −k

Qr2 ˆ r • d! r = −k

Qr2 dr.

∴V(r) = −k

Qr2 dr∫ = k

Qr

+ V",

where V" is an integration constant. If we define the

electric potential at infinity as zero, i.e., V(r → ∞) = 0, then V" = 0. So, the absolute electric potential at r is

V(r) = k

Qr

,

∴Vb − Va = kQ

1rb

−1ra

⎡

⎣ ⎢

⎤

⎦ ⎥ (i.e., < 0 if Q is +ve)

The electric field of a point charge is:

! E (r) = k

Qr2 ˆ r .+

Q

a b

• • d! r

r

The electric potential for a positive charge. If the charge is negative, the potential looks like a “hole” rather than a “hill”.

Note that as x (and y) → ±∞ , V→ 0.

V = k

Qr

Electricpotential (V)

x

y

“equipotentials”

Go from a → b by different routes. The potential at any point a distance r from a point charge is:

V(r) = k

Qr

.

Since ra = rx, Va = Vx, so

the potential difference

Vab = Vxb,

i.e., the potential difference between two points does not depend on the path between them only the potentials at the end points. The work done by you in moving a charge q from a → b by the two different routes is:

[1] Wa→b = −q(Va − Vb ) = −qVab.

[2] Wa→x→b = −q(Va − Vx )[ ]+ −q(Vx − Vb)[ ] = −q(Vx − Vb) = −qVxb.

But Vab = Vxb. ∴Wa→b = Wa→x→b.

So, the work done by you ( = −qΔV) in moving a charge from one point to another does not depend on the path ... only on ΔV.

b

a x

Question 23.4: Is the electric force a conservative or non-conservative force?

The electric force is a conservative force. Here are two reasons ...

[1] The work done by the electric force in moving a charge from one point to another is independent of the path ... a property of a conservative force.

[2] We can write a potential (energy) function, which can only be done for conservative forces.

For a point charge V(r) = k

Qr

. Therefore, the potential

energy of a test charge q! in the field due to a point charge Q is

U(r) = q!V(r) = k

q!Qr

.

i.e., a simple function of r.

No matter what the source, the electric force is a conservative force.

Question 23.5:

A proton is fired towards a helium nucleus. If the speed of the proton at point 1 is v1 and its speed at point 2 is

v2, which of the following statements is correct?

A: v2 < v1.

B: v2 > v1.

C: v2 = v1.

+

+

+

1

2

Use the conservation of mechanical energy:

K2 + U2 = K1 + U1.

The potential energy of the proton (with charge q) in the

field of the helium nucleus at 1 is U1 = qV1, and its

potential energy at 2 is U2 = qV2.

Since V = k

Qr

and r2 < r1, then V2 > V1.

∴U2 > U1.

So, K2 < K1,

i.e., v2 < v1,

which means A is the correct choice.

r2

r1

+

+

+

1

2

Question 23.6: Points A, B and C are at the vertices of an equilateral triangle whose sides are 3.00 m long. Point charges of +2.00 µC are fixed at A and B.

(a) What is the electric potential at point C?

(b) How much work is required to move a +5.00 µC point charge from infinity to the point C?

(c) How much additional work is required to move the

+5.00 µC charge from C to the midpoint of side AB?

(a) Potential at C is due to both QA and QB ( = Q)

VC = k

QAr

+ kQBr

= 2kQr

,

where r is the length of the sides. Remember V is a scalar.

∴VC = 2 × 9 ×109( ) × 2 ×10−6( )

3= 1.2 ×104 volts.

(b) Work done in bringing a charge q! from ∞ to the

point C is:

W = −q!ΔV = −q!(V∞ − VC) = q!VC

= 5 ×10−6( ) × 1.2 ×104( ) = 6.0 ×10−2J.

+2µC A

B C 3m 3m

3m

+2µC V = k

Qr

(c) The extra work done in moving the charge from C to D is

δWC→D = −q! VC − VD( ) = q!(VD − VC)

But VD = k

QAr2( ) + k

QBr2( ) = 4k

Qr

From earlier, VC = 2k

Qr

.

∴δWC→D = q! 4k

Qr− 2k

Qr

⎛ ⎝ ⎜

⎞ ⎠ ⎟ = 2q!k

Qr

= 2 × 5 ×10−6( ) × 9 ×109( ) × 2 ×10−6( )

3

= 6.0 ×10−2J.

+2µC A

B C +2µC

×D

From earlier: E = −

dVdr

so dV = −E.dr,

Potential due to a spherical shell of charge • on a hollow or solid conducting sphere ...

σ⇒

Q4πR2

But what about inside the sphere?

+ +++

+ +

+R+

E(r)

r E = 0

E = k

Qr2

Es = k

QR2 =

σε!

r

V = k

Qr

Vs = k

QR

V(r) ∴V(r > R) = − E.dr

r>R∫ = −k

Qr2∫ dr = k

Qr

Potential due to a spherical shell of charge • on a hollow or solid conducting sphere ...

E(r)

r E = 0

E = k

Qr2

Es = k

QR2 =

σε!

Inside the sphere, i.e., for r < R, dV = −E.dr = 0

∴V(r < R) = constant.If V is constant inside sphere, no work is done in moving a charge anywhere inside the sphere. Then

W = −q!ΔV = 0, i.e., ΔV = 0.

∴V(r < R) = k

QR⇒ constant.

r

V = k

Qr

Vs = k

QR

V(r)

TWO POINTS:[1] V is constant inside a conducting sphere

(i.e., the same as at the surface).

[2] At the surface: Vs = k

QR

and Es = k

QR2 .

∴Vs = Es.R or Es =

VsR

.

As the charge Q on the sphere increases, so do

Vs = kQ

R( ) and Es = kQ

R2⎛ ⎝ ⎜ ⎞

⎠ ⎟ .

Under “normal conditions” the maximum electric field obtainable in air before breakdown is

Emax ~ 3 ×106 V/m.

This sets a maximum potential and a maximum charge for a spherical conductor (radius R):

i.e., Vmax = Emax.R ~ 3 ×106R volts.

Since, E = k

QR2 , then

Qmax ~

3 ×106

kR2 Coulombs.

Larger R means larger Vmax and Qmax before breakdown.

Question 23.7: We know that the electric field near the Earth’s surface is ~200 V/m. If the Earth has a radius of about 6400 km, what is the Earth’s electric potential?

Assume the Earth is a sphere. The potential at the surface of a sphere is:

Vs = Es.R.

For the Earth: Vs = Es.R = 200 V/m( ) × (6400 ×103 m)

= 1.28 ×109 volts!

Note: in question 22.4, we found that the net charge on

the Earth is negative ... −9.11×105 C ... so the electric

potential at the surface is

Vs = k

QR

< 0

DISCUSSION PROBLEM:

If the electric potential of the Earth is so large, how come

we aren’t fried to a crisp when standing barefoot on the

Earth’s surface?

Also, we can now show why charges “pile-up” at sharp points on a charged conductor ...

The potential inside the conductor is constant

∴k

Q1R1

= kQ2R2

.

But Q1 ≈ 4πR12σ1 and Q2 ≈ 4πR2

2σ2

so R1σ1 ≈ R2σ2

i.e., σ2 ≈

R1R2

σ1. ∴σ2 > σ1.

Also, E1s =

σ1ε!

and E2s =

σ2ε!

. ∴E2s > E1s.

Therefore, the charge density and the surface electric field are greater at “points”.

Region 2 ⇒ Q2radius R2

Region 1 ⇒ Q1radius R1

σ2 σ1

1 2 3

Question 23.8: These objects are charged to their maximum potential (voltage) before breakdown of the surrounding air. Rank them in order from the one with the largest voltage to the smallest voltage.

We know that V = EsR. Therefore, the maximum

potential (when breakdown occurs) Vmax ∝R. Since

R3 > R1 > R2, the order from largest maximum potential

to smallest is:

3 :1: 2

Note that at the sharp end of #2 the electric field is greatest; that’s where breakdown first occurs.

1 2 3

R1

R2

R3 R3

The force on charge q is: F = qE = −q

dVdℓ

, but VA > VB

so dVdℓ

< 0, therefore, +ve charges move from A to B, i.e.,

from high potential ( VA) to low potential ( VB).

As charges move from A to B, VA decreases and VB

increases. When VA = VB, charges stop moving because

when ΔV = 0 then F = 0.

Take two charged spheres A ( VA) and B ( VB) with

VA > VB ... connect them together by a conducting wire.

A B

VA = k

QARA

VB = kQBRB

+ q

V ∴QA and VA decrease

∴QB and VB increase

+

Question 23.9: A spherical conductor of radius

R1 = 24.0 cm is charged to 20.0 kV. When it is

connected by a long conducting wire to a second conducting sphere a great distance away, its potential drops to 12.0 kV. If the second sphere was initially uncharged, what is the radius, R2, of the second sphere?

After the spheres are connected we have:

V1′ = k

q1R1

= V2′ = k

q2R2

= 12 ×103V,

i.e., q1 = (12 ×103)R1

k and

q2 = (12 ×103)R2

k... [1]

The total charge (q1 + q2) is conserved, before they are

connected we have

V1 = k

(q1 + q2)R1

= 20 ×103V,

i.e., q1 + q2 = 20 ×103( )R1

k.

Using q1 and q2 from [1],

(12 ×103)R1

k+ (12 ×103)R2

k= (20 ×103)R1

k,

i.e., (12 ×103)R2 = (8 ×103)R1.

∴R2 =

812

R1 =23× 24.0 cm = 16.0 cm.

R2 R1

1 2

V1 = 20kV

V1′ = 12kV

V2 = 0

V2′ = 12kV

Question 23.10: Initially, two, well separated conducting spheres are given the same charge (q). They are then connected to each other with a conducting wire. If

R2 = 2R1, what are the final charges, q1 and q2, and

charge densities, σ1 and σ2, on each sphere?

q1 q2

R2 = 2R1

R1 R2

q1 q2

R2 = 2R1

R1 R2

If they have the same initial charge, the potential of #1 ( V1) will be greater than the potential of #2 ( V2), because

R1 < R2. After connecting the two spheres together,

charges will flow from #1 to #2 until the potentials are equal, i.e., V1 = V2. Then

k

q1R1

= kq2R2

, i.e.,

q1R1

=q2

2R1.

∴q2 = 2q1.

The surface area 4πR2( ) of #2 is four times the surface

area of #1, i.e., A2 = 4A1.

∴σ2 =

q2A2

=2q14A1

=12σ1.

The potential at the point P due to the element of charge dQ a distance r from P is:

dV = k

dQr

.

∴V(x) = k

dQr∫ .

But r = x2 + a2 ⇒ constant.

∴V(x) = k

x2 + a2 dQ∫ = kQ

x2 + a2

where Q is the total charge on the ring.

Electrostatic potential due to a uniformly charged ring:

+

x

a

+

++

+

+ dQ

dV P

r = x2 + a2

×

Electrostatic potential due to a uniformly charged ring:

+

x

a

+

++

+

+ dQ

dV P

r = x2 + a2

×

• When x = 0, V(x) = k

Qa

. Also, Ex = −

dVdx

= 0.

• When x >> a, V(x)⇒ k

Qx

,

i.e., the ring looks like a point charge.

V(x) = k

Q

x2 + a2

-2 0 2 4 6

V = k

Qa

x

a

V = k

Qx

Question 23.11: What difference would it make, if any, if the charge Q were NOT uniformly nor symmetrically distributed around the ring?

A: The potential at any point P on the axis would be larger.

B: The potential at any point P on the axis would be smaller.

C: The potential depends on the actual distribution.D: It would make no difference.

x a P

×

Take an extreme scenario, i.e., with all the charge concentrated at one point on the ring. The total potential at P due to dQ is

V = k

dQr

⇒ kQr

,

i.e., the same as before!

Since potential is a scalar quantity it makes no difference if the charge is concentrated in one region or distributed around the ring; it only depends on r, the distance of P from the ring.

Therefore, the answer is D.

a

dQ⇒ Q

V

r P×

Three dimensional plot of the electric potential for two equal and opposite charges.

x y

0 0

“equipotentials” V

0

Question 23.12: Several equipotentials are shown labelled in volts. The spacing of the grid is 1.00cm. What is the magnitude and direction of the electric field at

(a) X, (b) Y?

1 cm

1 cm

X Y

0V 5V

10V

15V

20V x

y

The electric fields at X and Y are parallel to the x-axis.

∴! E = −

∂V∂x

ˆ i .

(a) at X:

! E = −

ΔVΔ! x

= −(15 V − 5 V)

0.01 mˆ i = −

10 V0.01 m

ˆ i

= −1000ˆ i V/m.

The negative sign means the electric field is in the −x

direction. (Note also that ! E is always directed from

higher potential to lower potential ... it’s the direction a

+ve charge would move.)

1 cm

1 cm

X Y

0V 5V

10V

15V

20V x

y

1 cm

1 cm

X Y

0V 5V

10V

15V

20V x

y

(b) at Y:

! E = −

ΔVΔ! x

= −(5V −15V)

0.02mˆ i = 10V

0.02mˆ i

= +500ˆ i V/m.

The positive sign means the electric field is in the +x direction.

Question 23.12: Point charges, q1 = q3 = −4.20 µC and

q2 = +4.20 µC are fixed at the vertices of an equilateral

triangle, whose sides are 2.50 m long. What is the electrostatic potential energy of this ensemble of charges?

+4.2µC

−4.2µC −4.2µC

2.50 m 2.50 m

2.50 m

The potential energy of an ensemble of charges is simply equal to the work done in bringing the charges together. Assume charge q1 is at position A and bring in charge q2

to B from ∞. The potential at B due to charge q1 at A is:

VB = k

q1rAB

.

So, the work done in bringing a charge q2 from ∞ to B is:

W2 = −q2ΔV = −q2(V∞ − VB) = −q2(0 − VB)

= k

q1q2r12

= (9 ×109) (−4.2 ×10−6)(4.2 ×10−6)2.50

= −6.35 ×10−2J.

q2 = +4.2µC

q1 = −4.2µC

2.50 m

A

B

Now bring in a charge q3 to the point C. The work done

in bringing charge q3 from ∞ to C is:

W3 = −q3(V∞ − VC ), where VC is the potential at C due to the charges at A

and B.

∴W3 = −q3 0 − (k q1

r13+ k

q2r23)⎛

⎝ ⎜ ⎞

⎠ ⎟ = k

q1q3r13

+ kq2q3r23

= (9 ×109)(−4.2 ×10−6)(−4.2 ×10−6)

2.50

+(9 ×109)(4.2 ×10−6)(−4.2 ×10−6)

2.50

= 0.So, the total work done

= W2 + W3 = −6.35 ×10−2J.

q3 = −4.2µC

q2 = +4.2µC

q1 = −4.2µC

2.50 m 2.50 m

2.50 m A

B

C

?

Electrostatic potential energy of a charged sphere

The electrostatic potential energy of a charged conducting sphere is equal to the amount of work we do in putting the

charge onto the sphere. If the sphere already has a charge q, the work done in bringing a charge dq from ∞

onto the sphere is dW = −dq(V∞ − V), where V is the

potential of the sphere. But V∞ = 0 and V = k q

R( ),

∴dW = k q

R( )dq.

The total work done to charge the sphere from 0→ Q is:

W = k q

R( )dq0

Q∫ =

kR

q2

2⎡ ⎣ ⎢

⎤ ⎦ ⎥ 0

Q

=12

kQ2

R =12

QV,

where V ( = kQR) is the potential of the fully charged

sphere.

∴U = W =

12

QV.

R dqq +

Question 23.13: An isolated spherical conductor with a radius of 10.0 cm is charged to 2.00 kV.

(a) What is the electrostatic potential energy of the sphere?

(b) What is the charge on the surface of the sphere?

(a) We’ve just shown that the electrostatic potential energy of a

charged sphere is: U =

12

QV.

But, V = k

QR

, i.e., Q =

VRk

.

Substituting for Q, we find

U =

12

VRk

⎛ ⎝

⎞ ⎠ V =

V2R2k

=(2.00 ×103)2 × 0.10

2 × 9 ×109 = 2.22 ×10−5J.

This is the amount of work we do in charging the sphere to a potential of 2.00 kV.

(b) The charge on the sphere is (from above):

Q =

RVk

=0.10 × 2.00 ×103

9 ×109 = 2.22 ×10−8C

= 22.2 nC.

RQ