Q109 / 150 · MHT CET 2023 (09 May Shift 2) · Physics

When a charge of $3 \mathrm{C}$ is placed in uniform electric field, it experiences a force of $3000 \mathrm{~N}$. Within this field, potential difference between two points separated by a distance of $1 \mathrm{~cm}$ is

  • Correct answer
    A
    $10 \mathrm{~V}$
  • B
    $90 \mathrm{~V}$
  • C
    $1000 \mathrm{~V}$
  • D
    $3000 \mathrm{~V}$

Solution

Electric force, $F=q E$ and potential difference, $\mathrm{V}=\mathrm{Ed}$ $\therefore \quad \mathrm{V}=\frac{\mathrm{Fd}}{\mathrm{q}}=\frac{3000 \times 10^{-2}}{3}=10 \mathrm{~V}$

Hence, the correct answer is option: 'A'.

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