Physics Electrostatics ETEA 2023
PMDC Verified Question 46 of 131
Q.163 The force of repulsion between two alike charges is 10N in vacuum. When a material of \( \epsilon_r = 2 \) is placed between them, new force will be:
A
20 N
B
15 N
C
10 N
D
5 N
Tap any option to test your recall and reveal the step-by-step Propolis autopsy.

Propolis Cognitive Error Autopsy

Official Correct Choice:
Option D: 5 N
1. Concept:

Inserting a dielectric material between two free charges shields them from each other, dampening the electrostatic interaction force between them.

2. Formula:

$$ F_{medium} = \frac{F_{vacuum}}{\epsilon_r} $$

3. Solution:

  • Original force in vacuum: \( F_{vac} = 10 \text{ N} \).


  • Dielectric constant of new medium: \( \epsilon_r = 2 \).


  • Substitute and divide: \( F_{med} = \frac{10}{2} = 5 \text{ N} \).


4. Why other options are incorrect:

Option A results from multiplying the force by the dielectric constant (which increases Capacitance, but always decreases Force).

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