+91-8076753736[email protected]
logo
logo
-->

Electrostatic Potential And Capacitance

Question
CBSEENPH12039967
Wired Faculty App

The electric potential V at any point (x, y,z), all in meters in space is given by V= 4x2 volt. The electric field at the point (1,0,2) in volt/meter is 

  • 8 along positive X-axis

  • 16 along negative x- axis

  • 16 along positive X- axis

  • 8 along negative X - axis

Solution

D.

8 along negative X - axis

We know that

straight E space equals space minus open square brackets bold i fraction numerator partial differential straight V over denominator partial differential straight x end fraction plus bold j fraction numerator partial differential straight V over denominator partial differential straight y end fraction plus bold space bold k fraction numerator partial differential straight V over denominator partial differential straight z end fraction close square brackets So comma space straight E space equals space minus space straight i space fraction numerator partial differential straight V over denominator partial differential straight x end fraction space equals space minus straight i fraction numerator partial differential left parenthesis 4 straight x right parenthesis squared over denominator partial differential straight x end fraction equals space minus 8 xi space Vm to the power of negative 1 end exponent straight E subscript left parenthesis 1 comma 0 comma 2 right parenthesis end subscript space equals space minus space 8 space bold i bold space Vm to the power of negative 1 end exponent

Some More Questions From Electrostatic Potential and Capacitance Chapter

Mock Test Series

Mock Test Series

NCERT Sample Papers

Entrance Exams Preparation