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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
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27 MV
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
$9 \mathrm{MV}+18 \mathrm{MV}=27 \mathrm{MV}$
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
$9 \mathrm{MV}+18 \mathrm{MV}=27 \mathrm{MV}$
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