Chapter 24: Q13P (page 710)

What are (a) the charge and (b) the charge density on the surface of a conducting sphere of radius 0.15 mwhose potential is 200 V(with V = 0at infinity)?

Short Answer

Expert verified

The charge is $q = 3.3 \times 10^{- 9} C$ .
The charge density on the surface of a conducting sphere is $σ = 1.2 \times 10^{- 8} C / m^{2}$ .

Step by step solution

Given data:

Radius of the conducting sphere, $R = 0.15 m$
Potential of the conducting sphere, $V = 200 V$

Understanding the concept:

Using the equation of potential energy and surface charge density, you find charge on the sphere and surface charge density respectively.

$V = \frac{k q}{R}$ ….. (1)

Here, V is the potential, q is the charge, R is the radius, k is the Coulomb’s constant having the value

of $8.99 \times 10^{9} N \cdot m^{2} / C^{2}$ .

(a) Calculate the charge on the surface of a conducting sphere

The charge on the sphere is define by rearrange equation (1) for the charge as below.

$\begin{array}{rcl} q & = & \frac{V R}{k} \\ = & \frac{(200 V) (0.15 m)}{8.99 \times 10^{9} N \cdot m^{2} / C^{2}} \\ = & 3.3 \times 10^{- 9} C \end{array}$

Hence, the charge is $\begin{array}{rcl} 3.3 \times 10^{- 9} C \end{array}$ .

(b) Calculate the charge density on the surface of a conducting sphere:

The (uniform) surface charge density (charge divided by the area of the sphere) is,

$\begin{array}{rcl} σ & = & \frac{q}{4 π R^{2}} \\ = & \frac{3.3 \times 10^{- 9} C}{4 \times 3.14 {(0.15 m)}^{2}} \\ = & 1.2 \times 10^{- 8} C / m^{2} \end{array}$