Physics Electrostatics ETEA 2018
PMDC Verified Question 111 of 131
Q.104 The force between two charged bodies is "F". if one of the charge is doubled and the distance between them is halved, the force acting on each charged body is:
A
2F
B
4F
C
8F
D
16F
Tap any option to test your recall and reveal the step-by-step Propolis autopsy.

Propolis Cognitive Error Autopsy

Official Correct Choice:
Option C: 8F
1. Concept:

Electrostatic force is directly proportional to the magnitude of the charges and inversely proportional to the square of the distance between them.

2. Formula:

$$ F_{new} = \frac{k (2q_1) q_2}{(r/2)^2} $$

3. Solution:

  • Original force: \( F = \frac{k q_1 q_2}{r^2} \).


  • Substitute new parameters: \( F_{new} = \frac{k (2q_1) q_2}{r^2 / 4} \).


  • Bring the denominator fraction up: \( F_{new} = 2 \times 4 \times \frac{k q_1 q_2}{r^2} \).


  • Result: \( F_{new} = 8 F \).


4. Why other options are incorrect:

Option A results from forgetting to halve the distance. Option B results from forgetting to double the charge. Option D assumes both charges were doubled.

Quality & Fidelity Assurance: Every question on BeambePrep is rigorously curated against the official PMDC syllabus with zero filler, zero out-of-syllabus content, and zero typos. When an authentic past paper originally contained a historical mistake or ambiguity from the examining board (such as UHS or NUMS), BeambePrep faithfully reflects the original paper while detailing the nuance and scientific consensus in the autopsy above.

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