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BITSAT 2021
Exam
BITSAT
Year
2021
Subject
Physics
Chapter
Electrostatics

BITSAT 2021 · Physics · Electrostatics

The electric potential V is given as a function of distance ×(...

The electric potential VV is given as a function of distance ×(\times( metre )) by V=V=\left(5 x^{2}+10 x-4\right)volt.<strong>Question:</strong>Valueofelectricfieldatvolt. <strong>Question:</strong> Value of electric field atx=1 \mathrm{ m} is

Options

  1. A−23V/m-23 \mathrm{ V} / \mathrm{m}
  2. B11V/m11 \mathrm{ V} / \mathrm{m}
  3. C6V/m6 \mathrm{ V} / \mathrm{m}
  4. D−20V/m-20 \mathrm{ V} / \mathrm{m}

Correct answer

D. −20V/m-20 \mathrm{ V} / \mathrm{m}

Step-by-step solution

V=5x2+10x−4E=−dVdx=−(10x+10)\mathrm{V}=5 x^{2}+10 x-4\mathrm{E}=\frac{-\mathrm{dV}}{\mathrm{dx}}=-(10 x+10) At x=1m,E=−20V/mx=1 m, \mathrm{E}=-20 \mathrm{ V} / m
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