Physics Electrostatics KU 2024
PMDC Verified Question 8 of 131
Q.185 What is the increase in force between two charges if the separation between them is decreased by 50 percent?
A
Becomes four times
B
Doubles
C
Increases by half
D
Triples
Tap any option to test your recall and reveal the step-by-step Propolis autopsy.

Propolis Cognitive Error Autopsy

Official Correct Choice:
Option A: Becomes four times
1. Concept:

Electrostatic force scales according to the inverse square law of the distance separating the two charges.

2. Formula:

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

3. Solution:

  • Decreasing the distance by 50 percent means the new distance is exactly half the original: \( r' = 0.5r = r/2 \).


  • Substitute this new distance into the denominator: \( F' \propto \frac{1}{(r/2)^2} \).


  • Square the term: \( F' \propto \frac{1}{r^2/4} \).


  • Invert the fraction: \( F' = 4 \left( \frac{1}{r^2} \right) \).


  • The final force is exactly 4 times the original force.


4. Why other options are incorrect:

Option B (doubling) occurs if you forget to square the distance change. Options C and D are not mathematically possible via the inverse square law for a 50% change.

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.

Want to solve full-length papers under timed exam conditions?

Practice with zero-scroll lockdown sprints, dynamic latency zone timers, live peer selection telemetry, and the automated Amber mistake recovery loop.