JEE Main 2016PhysicsCapacitanceHardMCQ

JEE Main 2016Capacitance Question with Solution

JEE Main 2016 (03 Apr)

Question

A combination of capacitors is set up as shown in the figure. The magnitude of the electric field, due to a point charge Q (having a charge equal to the sum of the charges on the 4 μF and 9 μF capacitors), at a point distant 30 m from it, would equal:
 

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Show full solutionCorrect option: A
Correct answer
A420 N/C

Step-by-step explanation



Ceq=5μF

Potential of 4 μF= 6 volt

charge on 4 μF q4=24 μC

Potential of 9 μF=2 volt

charge on 9 μF q9=18 μC
Total charge (q)=42 μC 
E=kqr2=9×109×42×10-6900=420 N/C

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About this question

This is a previous-year question from JEE Main 2016, covering the Capacitance chapter of Physics. PrepSharp catalogues every PYQ from JEE Main with a verified answer key and step-by-step solution prepared by IIT alumni — so you can search by chapter, topic or year and revise efficiently.