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KCET · Physics · Electrostatics

The magnitude of point charge due to which the electric field \( 30 \mathrm{~cm} \) away has the magnitude \( 2 \mathrm{~N} C^{-1} \) will be

  1. A \( 2 \times 10^{-11} \mathrm{C} \)
  2. B \( 3 \times 10^{-11} \mathrm{C} \)
  3. C \( 5 \times 10^{-11} \mathrm{C} \)
  4. D \( 9 \times 10^{-11} \mathrm{C} \)
Verified Solution

Answer & Solution

Correct Answer

(A) \( 2 \times 10^{-11} \mathrm{C} \)

Step-by-step Solution

Detailed explanation

Electric field due to a point charge is
\(E=\frac{1}{4 \Pi \varepsilon_{0}} \frac{q}{r^{2}}\)
Given,
\(E=2 N C^{-1} ; r=30 \mathrm{~cm}=30 \times 10^{-2} m ; \frac{1}{4 \Pi \varepsilon_{0}}=9 \times 10^{9} \)
\(\Rightarrow 2=9 \times 10^{9} \times \frac{q}{\left(30 \times 10^{-2}\right)^{2}} \)
\(\Rightarrow q=\frac{2 \times\left(30 \times 10^{-2}\right)^{2}}{9 \times 10^{9}}=2 \times 10^{-11} \mathrm{C}\)
Therefore, magnitude of charge will be \( 2 \times 10^{-11} \mathrm{C} \)