First slide

Coulumbs law

Question

The point charges $3 μ C and 4 μ C$ 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

Difficult

A

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

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

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

$0.18 \times 10^{- 3} N$

Solution

For finding air equivalent separation
We have $\frac{1}{4 π ϵ_{0}} \frac{q_{1} q_{2}}{r^{' 2}} = \frac{1}{4 π \in_{0} \in_{r}} \frac{q_{1} q_{2}}{r^{2}}$ $F^{'} = \frac{1}{4 π \in_{0}} \frac{q_{2} q_{2}}{r^{' 2}}$
$\Rightarrow r^{'} = \sqrt{ϵ_{r}} r$ $= 1.73 \times 10^{- 4} N$
$∴ r_{equivalent} = (\sqrt{9}) 3 + (\sqrt{16}) 4 = 25 m$
Therefore , F' = $\frac{1}{4 {πε}_{0}} \frac{q_{1} q_{2}}{{(r_{e q u i v a l e n t})}^{2}} = 9 \times 10^{9} \times \frac{3 \times 4 \times 10^{- 12}}{25^{2}} N = 1.732 \times 10^{- 4} N$

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