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Two point charges $A=+3 \mathrm{nC}$ and $B=+1 \mathrm{nC}$ are placed $5 \mathrm{~cm}$ apart in air. The work done to move charge $B$ towards $A$ by $1 \mathrm{~cm}$ is
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Verified Answer
The correct answer is:
$1.35 \times 10^{-7} \mathrm{~J}$
The given situation is shown below
$$
\begin{aligned}
r_{1} &=1 \mathrm{~cm} \\
&=1 \times 10^{-2} \mathrm{~m} \\
r_{2} &=5-1 \\
&=4 \mathrm{~cm} \\
&=4 \times 10^{-2} \mathrm{~m}
\end{aligned}
$$
Required work done $=$ Changes in potential energy
$$
\begin{aligned}
&=k q_{A} q_{B}\left[\frac{1}{r_{2}}-\frac{1}{r_{1}}\right] \\
&=9 \times 10^{9} \times 3 \times 10^{-9} \times 10^{-9} \\
&{\left[\frac{1}{4 \times 10^{-2}}-\frac{1}{1 \times 10^{-2}}\right]} \\
&=1.35 \times 10^{-7} \mathrm{~J}
\end{aligned}
$$
$$
\begin{aligned}
r_{1} &=1 \mathrm{~cm} \\
&=1 \times 10^{-2} \mathrm{~m} \\
r_{2} &=5-1 \\
&=4 \mathrm{~cm} \\
&=4 \times 10^{-2} \mathrm{~m}
\end{aligned}
$$
Required work done $=$ Changes in potential energy
$$
\begin{aligned}
&=k q_{A} q_{B}\left[\frac{1}{r_{2}}-\frac{1}{r_{1}}\right] \\
&=9 \times 10^{9} \times 3 \times 10^{-9} \times 10^{-9} \\
&{\left[\frac{1}{4 \times 10^{-2}}-\frac{1}{1 \times 10^{-2}}\right]} \\
&=1.35 \times 10^{-7} \mathrm{~J}
\end{aligned}
$$
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