Two charges 3 × 10^{-8} C and -2 × 10^{-8} C are located 15 cm apart. At what point on line joinin

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2. Electric Potential and Capacitance

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Two charges $3 \times 10^{-8}\; C$ and $-2 \times 10^{-8}\; C$ are located $15 \;cm$ apart. At what point on the line joining the two charges is the electric potential zero? Take the potential at infinity to be zero.

Option A

Option B

Option C

Option D

Solution

Let us take the origin $O$ at the location of the positive charge. The line joining the two charges is taken to be the $x-$axis; the negative charge is taken to be on the right side of the origin

Let $P$ be the required point on the $x$ -axis where the potential is zero.

If $x$ is the $x$ -coordinate of $P$, obviously $x$ must be positive. (There is no possibility of potentials due to the two charges adding up to zero for $x < 0 .$ If $x$ lies between $O$ and $A$, we have

$\frac{1}{4 \pi \varepsilon_{0}}\left[\frac{3 \times 10^{-8}}{x \times 10^{-2}}-\frac{2 \times 10^{-8}}{(15-x) \times 10^{-2}}\right]=0$

where $x$ is in $cm .$ That is,

$\frac{3}{x}-\frac{2}{15-x}=0$

which gives $x=9 cm$

If $x$ lies on the extended line $OA$, the required condition is $\frac{3}{x}-\frac{2}{x-15}=0$

which gives $x=45\; cm$

Thus, electric potential is zero at $9 cm$ and $45 cm$ away from the positive charge on the side of the negative charge. Note that the formula for potential used in the calculation required choosing potential to be zero at infinity.

Standard 12

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