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Q.

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

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a

$3 \times 10^{- 11} C$

b

$9 \times 10^{- 11} C$

c

$2 \times 10^{- 11} C$

d

$5 \times 10^{- 11} C$

answer is A.

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Detailed Solution

$\begin{array}{l} E = \frac{K q}{r^{2}} = \frac{1}{4 π ε_{0}} \frac{q}{r^{2}} \\ K = \frac{1}{4 π ε_{0}} = 9 \times 10^{9}, r = 30 c m \\ = 30 \times 10^{- 2} \\ E = 2 N C^{- 1} q = ? \end{array}$

$\begin{array}{l} 2 = \frac{9 \times 10^{9} \times q}{{(30 \times 10^{- 2})}^{2}} \\ \Rightarrow q = \frac{2 \times {(30 \times 10^{- 2})}^{2}}{9 \times 10^{9}} = 2 \times 10^{- 11} C \end{array}$

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The magnitude of point charge due to which the electric field 30cm away has the magnitude 2NC-1 will be