Question

Two point charges \(q_1\) and \(q_2\) are \(r\) cm apart in a vacuum with force \(F\). What is the force if they are placed in a medium with dielectric constant \(K = 5\) at distance \(r/5\)?

• A. \(F\)
• B. \(2F\)
• C. \(5F\)
• D. \(25F\)

Hint:

From Coulomb's law, \(F \propto \frac{1}{Kr^2}\). If distance decreases by a factor of \(5\), force increases by \(25\). Then divide by the dielectric constant \(K\).

Answer 1

The Correct Option is C

Concept: According to Coulomb's law, the force between two charges in vacuum is: \[ F = \frac{1}{4\pi\epsilon_0}\frac{q_1 q_2}{r^2} \] When the charges are placed in a medium with dielectric constant \(K\), the force becomes: \[ F = \frac{1}{4\pi\epsilon_0 K}\frac{q_1 q_2}{r^2} \] Thus, the force is inversely proportional to the dielectric constant and inversely proportional to the square of the distance.

Step 1: Write the expression for the new force. If the distance becomes \(r/5\) and the medium has dielectric constant \(K=5\): \[ F_{\text{new}} = \frac{1}{4\pi\epsilon_0 K}\frac{q_1 q_2}{(r/5)^2} \]

Step 2: Simplify the expression. \[ (r/5)^2 = \frac{r^2}{25} \] \[ F_{\text{new}} = \frac{1}{K}\times 25 \left(\frac{1}{4\pi\epsilon_0}\frac{q_1 q_2}{r^2}\right) \]

Step 3: Express in terms of the original force \(F\). \[ F_{\text{new}} = \frac{25}{5}F \] \[ F_{\text{new}} = 5F \]