Question

Two charges each of $100$ micro coulomb are separated in a medium of relative permittivity $2$ by a distance of $5cm$. The force between them is :

• A. $0.36\times {10}^{5}N$
• B. $3.6\times {10}^{5}N$
• C. $1.8\times {10}^{4}dyne$
• D. $1.8\times {10}^{4}N$

Answer 1

Answer: (D)

$1.8\times {10}^{4}N$

We know force between two charges in a medium separated by a distance 'r'

$F=\frac{\left(q_1\right)\left(q_2\right)}{4\pi \epsilon r^2}$

where, ${ϵ}_r=\frac{ϵ}{{ϵ}_o}$

given, ${ϵ}_r=2$ $⟹ϵ=2\times {ϵ}_o$

$F=\frac{(100\times {10}^{-6})(100\times {10}^{-6})\times 9\times {10}^9}{2\times (5\times {10}^{-2}{)}^2}$

$F=\frac{90}{2\times 25\times {10}^{-4}}$

$F=\frac{90\times 100\times {10}^2}{50}$

$F=1\cdot 8\times {10}^{4}N$