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