When a charge of $20 \mathrm{C}$ is taken from one point to another separated by a distance of $0.2…

When a charge of $20 \mathrm{C}$ is taken from one point to another separated by a distance of $0.2 \mathrm{~m}$, work of $2 \mathrm{~J}$ is required to be done. What is the potential difference between the two points?
  1. $2 \times 10^{-2} \mathrm{~V}$
  2. $4 \times 10^{-4} \mathrm{~V}$
  3. $8 \mathrm{~V}$
  4. $0.1 \mathrm{~V}$

Solution

Charge, $q=20 \mathrm{C}$ $ \begin{aligned} d & =0.2 \mathrm{~m} \\ W & =2 \mathrm{~J} \end{aligned} $ Potential difference, $V=\frac{W}{q}$ $ =\frac{2}{20}=0.1 \mathrm{~V} $

Asked in: AP EAMCET 2020 (22 Sep Shift 2)

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