Electrostatics Question 1

Question 1 - 24 January - Shift 1

If two charges $q_1$ and $q_2$ are separated with distance ’ $d$ ’ and placed in a medium of dielectric constant $K$. What will be the equivalent distance between charges in air for the same electrostatic force?

(1) $d \sqrt{k}$

(2) $k \sqrt{d}$

(3) $1.5 d \sqrt{k}$

(4) $2 d \sqrt{k}$

Show Answer

Answer: (1)

Solution:

Formula: Coulomb Force

$F=\frac{1}{(4 \pi \varepsilon_0)} \frac{q_1 q_2}{kd^{2}}$ (in medium)

$F _{\text{Air }}=\frac{1}{4 \pi \varepsilon_0} \frac{q_1 q_2}{d^{\prime 2}}$

$F=F _{\text{Air }}$

$\frac{q_1 q_2}{4 \pi \varepsilon_0 kd^{2}}=\frac{q_1 q_2}{4 \pi \varepsilon_0 d^{\prime 2}}$

$d^{\prime}=d \sqrt{k}$