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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
Options:
\( 2 \mathrm{~N} C^{-1} \) will be
Solution:
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Verified Answer
The correct answer is:
\( 2 \times 10^{-11} \mathrm{C} \)
Electric field due to a point charge is
\[
E=\frac{1}{4 \Pi \varepsilon_{0}} \frac{q}{r^{2}}
\]
Given,
\[
\begin{array}{l}
E=2 N C^{-1} ; r=30 \mathrm{~cm}=30 \times 10^{-2} m ; \frac{1}{4 \Pi \varepsilon_{0}}=9 \times 10^{9} \\
\Rightarrow 2=9 \times 10^{9} \times \frac{q}{\left(30 \times 10^{-2}\right)^{2}} \\
\Rightarrow q=\frac{2 \times\left(30 \times 10^{-2}\right)^{2}}{9 \times 10^{9}}=2 \times 10^{-11} \mathrm{C}
\end{array}
\]
Therefore, magnitude of charge will be \( 2 \times 10^{-11} \mathrm{C} \)
\[
E=\frac{1}{4 \Pi \varepsilon_{0}} \frac{q}{r^{2}}
\]
Given,
\[
\begin{array}{l}
E=2 N C^{-1} ; r=30 \mathrm{~cm}=30 \times 10^{-2} m ; \frac{1}{4 \Pi \varepsilon_{0}}=9 \times 10^{9} \\
\Rightarrow 2=9 \times 10^{9} \times \frac{q}{\left(30 \times 10^{-2}\right)^{2}} \\
\Rightarrow q=\frac{2 \times\left(30 \times 10^{-2}\right)^{2}}{9 \times 10^{9}}=2 \times 10^{-11} \mathrm{C}
\end{array}
\]
Therefore, magnitude of charge will be \( 2 \times 10^{-11} \mathrm{C} \)
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