Question

A force of 80 newton acts between two point charges placed in air at a fixed distance. When the same charges are placed at the same distance in a dielectric medium, the force becomes 8 newton. The dielectric constant of the medium is:

• A. 10
• B. zero
• C. 0.1
• D. 8

Hint:

In a medium the Coulomb force drops by the factor of the dielectric constant, so \( K = F_{air}/F_{medium} \).

Answer 1

The Correct Option is A

Step 1: State the concept.
By Coulomb's law, the force between two point charges separated by a fixed distance \( r \) in a medium is \[ F_{med} = \frac{1}{4\pi\varepsilon_0 K}\cdot\frac{q_1 q_2}{r^2}, \] where \( K \) is the dielectric constant of the medium. In air (taken as vacuum), \( K = 1 \), so \[ F_{air} = \frac{1}{4\pi\varepsilon_0}\cdot\frac{q_1 q_2}{r^2}. \]

Step 2: Form the ratio.
Since the charges and the separation are unchanged, dividing the two expressions gives \[ \frac{F_{air}}{F_{med}} = K. \]

Step 3: Substitute the given values.
Here \( F_{air} = 80 \) N and \( F_{med} = 8 \) N, so \[ K = \frac{80}{8}. \]

Step 4: Arithmetic.
\[ K = 10. \]

Step 5: Why the other options are wrong.
A dielectric always weakens the force, so \( K > 1 \); this rules out 0.1 and zero. The value 8 wrongly quotes the medium force rather than the ratio. Hence option (i) is correct.
\[\boxed{K = 10}\]