Question

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

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

Step-by-step solution

Given data:

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

Understanding the concept:

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

$\mathit{V}\mathbf{=}\frac{\mathbf{k}\mathbf{q}}{\mathbf{R}}$ ….. (1)

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

of $\mathbf{8}\mathbf{.}\mathbf{99}\mathbf{\times }{\mathbf{10}}^{\mathbf{9}}\mathbf{}\mathit{N}\mathbf{·}{\mathit{m}}^{\mathbf{2}}\mathbf{/}{\mathit{C}}^{\mathbf{2}}$.

(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.

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

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,

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

Hence, the charge density on the surface of a conducting sphere is $\begin{array}{rcl}& & 1.2\times {10}^{-8}C/m^2\end{array}$.