25.6.2 The Electric Potential of a Charged Sphere

In practice, you are more likely to work with a charged sphere, of radius R and total charge Q, than with a point charge. Outside a uniformly charged sphere, the electric potential is identical to that of a point charge Q at the center. That is,

V_{sphere} = \frac{1}{4 π ϵ_{0}} \frac{Q}{r} (sphere of charge, r \geq R)

(25.31)

We can cast this result in a more useful form. It is customary to speak of charging an electrode, such as a sphere, “to” a certain potential, as in “Bob charged the sphere to a potential of 3000 volts.” This potential, which we will call $V_{0}$ , is the potential right on the surface of the sphere. We can see from Equation 25.31 that

V_{0} = V (at r = R) = \frac{Q}{4 π ϵ_{0} R}

(25.32)

Consequently, a sphere of radius R that is charged to potential $V_{0}$ has total charge

Q = 4 π ϵ_{0} R V_{0}

(25.33)

If we substitute this expression for Q into Equation 25.31, we can write the potential outside a sphere that is charged to potential $V_{0}$ as

V_{sphere} = \frac{R}{r} V_{0} (sphere charged to potential V_{0})

(25.34)

Equation 25.34 tells us that the potential of a sphere is $V_{0}$ on the surface and decreases inversely with the distance. The potential at $r = 3 R$ is $\frac{1}{3} V_{0}$ .

A photograph of a man touching a plasma ball with his fingers. — A *plasma ball* consists of a small metal ball charged to a potential of about 2000 V inside a hollow glass sphere. The electric field of the high-voltage ball is sufficient to cause a gas breakdown at this pressure, creating “lightning bolts” between the ball and the glass sphere.

EXAMPLE 25.8 A proton and a charged sphere

A proton is released from rest at the surface of a 1.0-cm-diameter sphere that has been charged to $+ 1000 V$ .

What is the charge of the sphere?
What is the proton’s speed at 1.0 cm from the sphere?

MODEL Energy is conserved. The potential outside the charged sphere is the same as the potential of a point charge at the center.

VISUALIZE FIGURE 25.27 shows the situation.

The figure illustrates a before-and-after pictorial representation of proton separated from a sphere.

FIGURE 25.27 Full Alternative Text

SOLVE

The charge of the sphere is

$Q = 4 π ϵ_{0} R V_{0} = 0.56 \times 10^{- 9} C = 0.56 nC$
A sphere charged to $V_{0} = + 1000 V$ is positively charged. The proton will be repelled by this charge and move away from the sphere. The conservation of energy equation $K_{f} + e V_{f} = K_{i} + e V_{i}$ , with Equation 25.34 for the potential of a sphere, is

$\frac{1}{2} m v_{f}^{2} + \frac{e R}{r_{f}} V_{0} = \frac{1}{2} m v_{i}^{2} + \frac{e R}{r_{i}} V_{0}$

The proton starts from the surface of the sphere, $r_{i} = R$ , with $v_{i} = 0$ . When the proton is 1.0 cm from the surface of the sphere, it has $r_{f} = 1.0 cm + R = 1.5 cm$ . Using these, we can solve for $v_{f} :$

$v_{f} = \sqrt{\frac{2 e V_{0}}{m} (1 - \frac{R}{r_{f}})} = 3.6 \times 10^{5} m/s$

REVIEW This example illustrates how the ideas of electric potential and potential energy work together, yet they are not the same thing.