Q19.3-24PE
Question
A \({\bf{0}}.{\bf{500}}{\rm{ }}{\bf{cm}}\) diameter plastic sphere, used in a static electricity demonstration, has a uniformly distributed \({\bf{40}}.{\bf{0}}{\rm{ }}{\bf{pC}}\) charge on its surface. What is the potential near its surface?
Step-by-Step Solution
VerifiedValue of the electric potential near the surface of the sphere is \(V = 144\;V\).
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 \(\begin{array}{c}V = \frac{{{U_{Qq}}}}{q}\\ = \frac{{kQ}}{r}{\rm{ }}...(1)\end{array}\) 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).
Here diameter of the plastic sphere is:
\(\begin{array}{c}D = (0.500\;cm)\left( {\frac{{1\;m}}{{100\;cm}}} \right)\\D = 0.00500\;cm\end{array}\)
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:
\(\begin{array}{c}q = (40.0{\rm{ }}pC)\left( {\frac{{1{\rm{ }}C}}{{{{10}^{12}}{\rm{ }}pC}}} \right)\\q = 40.0 \times {10^{ - 12}}{\rm{ }}C\end{array}\)
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:
\(\begin{array}{c}V = \frac{{\left( {8.99 \times {{10}^9}\;N \times {m^2}/C} \right)\left( {40.0 \times {{10}^{ - 12}}{\rm{ }}C} \right)}}{{2.50 \times {{10}^{ - 3}}\;m}}\\V = 144\;V\end{array}\)
Therefore, the electric potential value is obtained as \(144\;V\).