Physics Electrostatics BUMHS 2022
PMDC Verified Question 75 of 131
Q.135 The electric field intensity at a point situated 4 m from a point charge is 200 N/C. If the distance is reduced to 2 m, what will be field intensity?
A
400 N/C
B
800 N/C
C
600 N/C
D
1200 N/C
Tap any option to test your recall and reveal the step-by-step Propolis autopsy.

Propolis Cognitive Error Autopsy

Official Correct Choice:
Option B: 800 N/C
1. Concept:

The electric field of a point charge obeys the inverse square law. Halving the distance from the source dramatically magnifies the field strength.

2. Formula:

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

3. Solution:

  • Original state: \( E_1 = 200 \text{ N/C} \) at \( r_1 = 4 \text{ m} \).


  • New state: \( r_2 = 2 \text{ m} \), which means the distance is halved (\( r_2 = r_1 / 2 \)).


  • Because \( E \) varies inversely with the square of the distance, replacing \( r \) with \( r/2 \) introduces a factor of \( 1 / (1/2)^2 = 1 / (1/4) = 4 \).


  • The new field is 4 times stronger: \( 200 \times 4 = 800 \text{ N/C} \).


4. Why other options are incorrect:

Option A incorrectly assumes a linear relationship (distance halved = field doubled). Options C and D are mathematically unrelated to the inverse-square geometric scaling.

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