Question

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?

• A. \(d[k]^{1/2}\)
• B. \(k[d]^{1/2}\)
• C. \(1.5d[k]^{1/2}\)
• D. \(2d[k]^{1/2}\)

Hint:

Force in a medium is \(F=\frac{q_1q_2}{4\pi\varepsilon_0Kd^2}\).

Answer 1

The Correct Option is A

Step 1: Plan:
Think of the dielectric as making the charges look farther apart.

Step 2: Steps:
The force in the medium drops by $K$, the same as increasing the distance in air by a factor $\sqrt K$, since force varies as $1/d^2$. So the equivalent air distance is $d\sqrt K$.

Final Answer:
The equivalent distance is $d\sqrt K$, option (A). \[ \boxed{d\,k^{1/2}} \]