A spherical conductor of radius \( 2 \mathrm{~cm} \) is uniformly charged with \( 3 \mathrm{nC} \). What is the electric field
at a distance of \( 3 \mathrm{~cm} \) from the center of the sphere?

• A

  \( 3 \times 10^{6} \mathrm{Vm}^{-1} \)
• B

  \( 3 \mathrm{Vm}^{-1} \)
• Correct answer
  C

  \( 3 \times 10^{4} \mathrm{Vm}^{-1} \)
• D

  \( 3 \times 10^{-4} \mathrm{Vm}^{-1} \)

Solution

Electric field at a point at distance ' \( r \) ' is given as \( E=\frac{1}{4 \Pi \varepsilon_{0}} \frac{q}{r^{2}} \)

Here, \( q=3 \mathrm{nC}=3 \times 10^{-9} \mathrm{C} ; r=3 \mathrm{~cm}=3 \times 10^{-2} \mathrm{~m} \)
Therefore,
\( E=\left(9 \times 10^{\circ}\right) \times \frac{3 \times 10^{-9}}{\left(3 \times 10^{-2}\right)^{2}}=\frac{9 \times 10^{9} \times 3 \times 10^{-9}}{9 \times 10^{-4}} \) \( =3 \times 10^{4} \mathrm{~V} \mathrm{~m}^{-1} \)
Thus, electric field at a distance of \( 3 \mathrm{~cm} \) from center of the sphere carrying charge \( 3 \mathrm{nC} \) is \( 3 \times 10^{4} \mathrm{Vm}^{-1} \)

Hence, the correct answer is option: 'C'.