Q108 / 180 · JIPMER 2015 · Physics

Two charges of $+10 \mu \mathrm{C}$ and $+20 \mu \mathrm{C}$ are separated by a distance 2 cm . The net potential (electric) due to the pair at the middle point of the line joining the two changes, is

  • Correct answer
    A
    27 MV
  • B
    18 MV
  • C
    20 MV
  • D
    23 MV

Solution

Using the equation $V=\frac{Q}{4 \pi \varepsilon_0 I}$ The potential due to $+10 \mu \mathrm{C}$ is $V_1=\frac{\left(10 \times 10^{-6}\right) \times\left(9 \times 10^9\right)}{1 \times 10^{-2}}=9 \mathrm{MV}$ The potential due to $+20 \mu \mathrm{C}$ is $V_2=\frac{20 \times 10^{-6} \times 9 \times 10^9}{1 \times 10^{-2}}=18 \mathrm{MV}$ The net potential at the given point is $9 \mathrm{MV}+18 \mathrm{MV}=27 \mathrm{MV}$

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

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