Question

The electrostatic force between two charges kept in air is F. If 30% of the space between the charges is filled with a medium, then the electrostatic force between the charges becomes $\frac{F}{2.56}$. The dielectric constant of the medium is

• A. 8
• B. 3
• C. 9
• D. 4

Hint:

When a medium of thickness $t$ and dielectric constant $K$ is introduced between two charges separated by distance $r$, the effective distance in the denominator of the force formula becomes $r_{eff} = (r-t) + t\sqrt{K}$. This is the key to solving such problems.

Answer 1

The Correct Option is D

Step 1: State the formula for the force between two charges with an intervening medium.
The force between two point charges $q_1$ and $q_2$ separated by a distance $r$ is $F = \frac{1}{4\pi\epsilon} \frac{q_1 q_2}{r^2}$. When a medium of dielectric constant $K$ fills a thickness $t$ between the charges (with a total distance $r$), the effective distance in the denominator of the force is changed from $r$ to $r_{eff}$. \[ F' = \frac{1}{4\pi\epsilon_0} \frac{q_1 q_2}{[r_{eff}]^2}. \] The effective distance for the force calculation is given by: \[ r_{eff} = (r-t) + t\sqrt{K}. \] The force can be written as $F' = F_{air} \frac{r^2}{[r_{eff}]^2}$.
Step 2: Apply the given condition to find the effective distance.
The initial force in air is $F_{air} = F$. The final force is $F' = \frac{F}{2.56}$. The distance $r$ remains the same. The ratio of the forces is $\frac{F_{air}}{F'} = \frac{[r_{eff}]^2}{r^2} = 2.56$. \[ \frac{r_{eff}}{r} = \sqrt{2.56} = 1.6. \] So, $r_{eff} = 1.6r$.
Step 3: Use the given thickness of the medium.
The thickness of the medium is $t = 30%$ of the total distance $r$. \[ t = 0.3r. \]
Step 4: Substitute the effective distance formula and solve for K.
We have $r_{eff} = 1.6r$ and $t = 0.3r$. Substitute these into the effective distance formula: \[ r_{eff} = (r-t) + t\sqrt{K}. \] \[ 1.6r = (r - 0.3r) + 0.3r\sqrt{K}. \] \[ 1.6r = 0.7r + 0.3r\sqrt{K}. \] Divide by $r$: \[ 1.6 = 0.7 + 0.3\sqrt{K}. \] \[ 0.3\sqrt{K} = 1.6 - 0.7 = 0.9. \] \[ \sqrt{K} = \frac{0.9}{0.3} = 3. \] Squaring both sides gives the dielectric constant: \[ \boxed{K = 9}. \]