The point charges 3μC and 4μC are placed at a separation of 7 m. The medium between them is of two types as shown in figure , the electric force acting between them is

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$3.2 \times 10^{- 4} N$

$0.18 \times 10^{- 4} N$

$1.75 \times 10^{- 3} N$

$1.73 \times 10^{- 4} N$

answer is B.

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Detailed Solution

For finding air equivalent separation
We have

$\frac{1}{4 {πε}_{0}} \frac{q_{1} q_{2}}{r^{' 2}} = \frac{1}{4 {πε}_{0} ε_{r}} \frac{q_{1} q_{2}}{r^{2}}$

$r_{equivalent} = \sqrt{16} (4) + \sqrt{9} (3) = 25 m$

$F = 9 \times 10^{9} \frac{12 \times 10^{- 12}}{{(25)}^{2}} = 1.73 \times 10^{- 4} N$

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