91Ó°ÊÓ

The magnitude of electric field outside the surface of each sphere is\(E = 1 \times {10^6}\;{\rm{N}}/{\rm{C}}\)

The diameter of each sphere is\(d = 5\;{\rm{m}}\)

The distance for electric field from the centre of each sphere is \(D = 25\;{\rm{m}}\)

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 each sphere is given below:

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

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.

\(\begin{aligned}1 \times {10^6}\;{\rm{N}}/{\rm{C}} = \frac{{\left( {9 \times {{10}^9}\;{\rm{N}} \cdot {{\rm{m}}^2}/{{\rm{C}}^2}} \right)Q}}{{{{\left( {25\;{\rm{m}}} \right)}^2}}}\\Q = 0.0694\;{\rm{C}}\end{aligned}\)

The electric field outside the surface of each sphere is given as:

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

Substitute all the values in the above equation.

\(\begin{aligned}\vec E = \frac{{\left( {9 \times {{10}^9}\;{\rm{N}} \cdot {{\rm{m}}^2}/{{\rm{C}}^2}} \right)\left( {0.0694\;{\rm{C}}} \right)}}{{{{\left( {\frac{{5\;{\rm{m}}}}{2}} \right)}^2}}}\\\vec E = 0.0999 \times {10^9}\;{\rm{N}}/{\rm{C}}\\\vec E \approx {10^8}\;{\rm{N}}/{\rm{C}}\end{aligned}\)

Therefore, the electric field outside the surface of each sphere is \({10^8}\;{\rm{N}}/{\rm{C}}\) and the option (d) is correct.