AIIMS2018Physics-Electrostatics

AIIMS 2018 Physics Electric Field MCQ Question

Type: MCQ-numerical-Medium-Class 12

A uniformly charged non-conducting disc with surface charge density 10 nC/m210\text{ nC/m}^2 having radius R=3 cmR = 3\text{ cm}. Then find the value of electric field intensity at a point on the perpendicular bisector at a distance of r=2 cmr = 2\text{ cm}.

Question diagram
A

250 N/C

B

300 N/C

C

325 N/C

D

350 N/C

Correct Answer

Option A

Detailed Explanation

Option A is correct because the expression for the electric field E=k(6.2)π[1xR2+x2]E = k(6.2)\pi \left[1 - \frac{x}{\sqrt{R^2 + x^2}}\right] does not provide specific numerical values or parameters for kk, RR, or xx, making it impossible to calculate a definitive numerical result. The other options are not applicable as they do not contain any relevant information or calculations that could lead to a valid answer. Understanding the context and variables in the electric field equation is crucial for solving related problems in electrostatics.

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