Physics Electrostatics MDCAT 2019
PMDC Verified Question 103 of 131
Q.107 Electric field strength at a point between oppositely charge plates is E. If the distance between plates is reduced to half, what will be the new value of electric intensity?
A
4E
B
E/2
C
E/4
D
2E
Tap any option to test your recall and reveal the step-by-step Propolis autopsy.

Propolis Cognitive Error Autopsy

Official Correct Choice:
Option D: 2E
1. Concept:

For parallel plates held at a constant potential difference, the uniform electric field is directly proportional to voltage and inversely proportional to plate separation.

2. Formula:

$$ E = \frac{V}{d} $$

3. Solution:

  • Let the original distance be \( d \) and the original field be \( E \).


  • The new distance is \( d' = d / 2 \).


  • Substitute into the formula: \( E_{new} = \frac{V}{d/2} = 2 \left(\frac{V}{d}\right) = 2E \).


4. Why other options are incorrect:

Option B results from a direct proportional misinterpretation (thinking decreasing distance decreases the field). Options A and C confuse uniform field principles with point-charge inverse-square laws.

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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