Physics Electrostatics BUMHS 2022
PMDC Verified Question 77 of 131
Q.137 When the distance between two charged particles is halved, the Coulomb force between them becomes
A
One fourth of the initial force
B
Four times the initial force
C
Double the initial force
D
One half of the initial force
Tap any option to test your recall and reveal the step-by-step Propolis autopsy.

Propolis Cognitive Error Autopsy

Official Correct Choice:
Option B: Four times the initial force
1. Concept:

The electrostatic force between two charges is intensely sensitive to distance, governed by the inverse square law.

2. Formula:

$$ F \propto \frac{1}{r^2} $$

3. Solution:

  • Let the original force be \( F \) at distance \( r \).


  • If the new distance is \( r' = r / 2 \), substitute this into the proportion.


  • \( F' \propto \frac{1}{(r/2)^2} = \frac{1}{r^2/4} = 4 \left( \frac{1}{r^2} \right) \).


  • The resulting force is exactly four times the original.


4. Why other options are incorrect:

Option A occurs if the distance was doubled. Option C implies an inversely linear law, which violates Coulomb's theorem.

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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