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Manipal MET2017PhysicsElectrostatics

The electric potential is +100 V at a distance of 10 cm from a point charge q . Then, q is equal to

Options

  1. A+1.1 10⁻⁹ C
  2. B+1.1 10⁻³ C
  3. C3 C
  4. D3 10⁻⁵ C

Correct answer

A. +1.1 10⁻⁹ C

Step-by-step solution

V= 1 4 ₀ q r Given, 1 4 ₀ =9 10^9, V=+100 volt, r=10 ~cm =0.10 ~m 100= (9 10^9 ) 9 0.10 q= 100 0.10 9 10^9 =1.1 10⁻⁹ C

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