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Electric field intensity and electric potential at a certain distance from a point charge is 32 N/C and 16 J/C. What is the distance from the charge?(a) 50 m(b) 0.5 m(c) 10 m(d) 7 mThis question was posed to me in a national level competition.This question is from Electrostatic Potential due to a Point Charge in chapter Electrostatic Potential and Capacitance of Physics – Class 12

Answer»

Correct choice is (b) 0.5 m

For explanation I WOULD SAY: Electric field DUE to a charge q at a distance r is \(\frac {q}{kr^2}\)

 and electric potential at that point is \(\frac {q}{kr}\). Therefore, \(\frac {q}{kr^2}\)=32 and \(\frac {q}{kr}\)=16. Dividing these TWO EQUATIONS, we get r=0.5m. If the medium is air, k=1. Thus we can get the value of q=0.89 C.



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