Question

A $\mathbf{0}.\mathbf{500}\mathbf{c}\mathbf{m}$ diameter plastic sphere, used in a static electricity demonstration, has a uniformly distributed $\mathbf{40}.\mathbf{0}\mathbf{p}\mathbf{C}$charge on its surface. What is the potential near its surface?

Short answer

Value of the electric potential near the surface of the sphere is $V=144V$.

Step-by-step solution

Definition of electric potential due to single charge.

Electric Potential Due to a Single Charge: Place a test charge at the desired location,$q$, and calculate the electric potential energy$U_{Qq}$of the system containing the test charge and the source charge that generates the field.

There is an electric potential of $\begin{array}{c}V=\frac{U_{Qq}}{q}\\ =\frac{kQ}{r}...(1)\end{array}$there.

Where$k=8.99\times {10}^{9}N\cdot m^2/C$, is the constant of Coulomb.

The unit of electric potential is joule/coulomb and is called the volt(V).

Data given for the electric potential on the surface of sphere.

Here diameter of the plastic sphere is:

$\begin{array}{c}D=(0.500cm)\left(\frac{1m}{100cm}\right)\\ D=0.00500cm\end{array}$

And the radius of the sphere is then:

$\begin{array}{c}r=\frac{D}{2}\\ r=\frac{0.00500}{2}\\ r=2.50\times {10}^{-3}m\end{array}$

The charge on the surface of the sphere is:

$\begin{array}{c}q=(40.0pC)\left(\frac{1C}{{10}^{12}pC}\right)\\ q=40.0\times {10}^{-12}C\end{array}$

Calculation of the electric potential near the surface of sphere.

The electric potential near the surface of the sphere is found from equation:

$V=\frac{kq}{r}$

Now entering the values for$k$,$q$, and$r$, it gives:

$\begin{array}{c}V=\frac{\left(8.99\times {10}^{9}N\times m^2/C\right)\left(40.0\times {10}^{-12}C\right)}{2.50\times {10}^{-3}m}\\ V=144V\end{array}$

Therefore, the electric potential value is obtained as $144V$.