Question

Two point charges placed at a certain distance \(r\) in air exert force \(F\) on each other. Then the distance \(r_{1}\) at which these charges will exert the same force in a medium of dielectric constant \(K\) is

• A. \(r\)
• B. \(\frac{r}{K}\)
• C. \(\frac{r}{\sqrt{K}}\)
• D. \(r\sqrt{K}\)

Hint:

The force in a medium is reduced by the dielectric constant K, so equal force needs a smaller distance.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept
In air \(F=\dfrac{1}{4\pi\epsilon_0}\dfrac{q_1q_2}{r^2}\). In a medium of dielectric constant \(K\), \(F'=\dfrac{1}{4\pi\epsilon_0}\dfrac{q_1q_2}{Kr_1^2}\).

Step 2: Equate forces
\(\dfrac{1}{r^2}=\dfrac{1}{Kr_1^2}\).

Step 3: Solve
\(r_1^2=\dfrac{r^2}{K}\), so \(r_1=\dfrac{r}{\sqrt{K}}\).

Step 4: Compare the options
This is option (C).

Final Answer:
The distance is r divided by root K. \[ \boxed{\text{(C) }\dfrac{r}{\sqrt{K}}} \]