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
- A \( 2 \times 10^{-11} \mathrm{C} \)
- B \( 3 \times 10^{-11} \mathrm{C} \)
- C \( 5 \times 10^{-11} \mathrm{C} \)
- D \( 9 \times 10^{-11} \mathrm{C} \)
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 \( 2 \times 10^{-11} \mathrm{C} \)
\(E=\frac{1}{4 \Pi \varepsilon_{0}} \frac{q}{r^{2}}\)
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 \( 2 \times 10^{-11} \mathrm{C} \)
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