(a) How many excess electrons must be distributed uniformly within the volume of an isolated plastic sphere 30.0 cm in diameter to produce an electric field of magnitude 1390 N/C just outside the surface of the sphere? (b) What is the electric field at a point 10.0 cm outside the surface of the sphere?

The magnitude of electric field outside the surface of plastic sphere is\(E = 1390\;{\rm{N}}/{\rm{C}}\)

The diameter of plastic sphere is\(D = 30\;{\rm{cm}}\)

The distance from the surface of plastic sphere is \(d = 10\;{\rm{cm}}\)

The number of excess electron is found by calculating the charge due to electric field on the outside of sphere.

The electric field just outside the surface of plastic sphere is given below:

\(E = \frac{{kQ}}{{{{\left( {\frac{D}{2}} \right)}^2}}}\)

Here,\(k\)is Coulomb’s constant and its value is\(9 \times {10^9}\;{\rm{N}} \cdot {{\rm{m}}^2}/{{\rm{C}}^2}\).

Substitute all the values in the above equation.

\(\begin{aligned}1390\;{\rm{N}}/{\rm{C}} = \frac{{\left( {9 \times {{10}^9}\;{\rm{N}} \cdot {{\rm{m}}^2}/{{\rm{C}}^2}} \right)Q}}{{{{\left( {\left( {\frac{{30\;{\rm{cm}}}}{2}} \right)\left( {\frac{{1\;{\rm{m}}}}{{100\;{\rm{cm}}}}} \right)} \right)}^2}}}\\Q = 3.5 \times {10^{ - 9}}\;{\rm{C}}\end{aligned}\)

The number of excess electron on outside surface of plastic sphere is given as:

\(Q = ne\)

Here,\(e\)is the charge of an electron and its value is\(1.6 \times {10^{ - 19}}\;{\rm{C}}\).

Substitute all the values in the above equation.

\(\begin{aligned}3.5 \times {10^{ - 9}}\;{\rm{C}} = n\left( {1.6 \times {{10}^{ - 19}}\;{\rm{C}}} \right)\\n = 2.2 \times {10^{10}}\end{aligned}\)

Therefore, the number of excess electron on outside surface of plastic sphere is \(2.2 \times {10^{10}}\).