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: Understanding the Concept:
A dielectric of constant \(K\) reduces the force between two charges by a factor of \(K\).

Step 2: Equate:
In the medium: \(F = \frac{1}{4\pi\varepsilon_0}\frac{q_1q_2}{Kd^2}\). In air at distance \(d'\): \(F = \frac{1}{4\pi\varepsilon_0}\frac{q_1q_2}{d'^2}\).
Equal forces mean \(d'^2 = Kd^2\), so \(d' = d\sqrt K = d[K]^{1/2}\).

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