Question

Three point charges of $3 \,\mu C\,, 4 \,\mu C$, and $5 \,\mu C$ are arranged at the three corners of a right angled triangle $A B C$ as shown in the figure. The work done in moving the charges at $A$ and $C$, so that the three charges are located at the three corners of an equilateral triangle of side $3 \,cm$ is

• A. 0.3 J
• B. 1.1 J
• C. 2.2 J
• D. 3.3 J

Answer 1

Answer: (D)

According to question

In a right angle triangle, in $\Delta A B C$
$ A C^{2}=A B^{2}+B C^{2} $
$\Rightarrow A C=\sqrt{A B^{2}+B C^{2}}=\sqrt{\left(4 \times 10^{-2}\right)^{2}+\left(3 \times 10^{-2}\right)^{2}}$
$\Rightarrow A C=5 \times 10^{-2} m $
Initial electric potential energy of three charges,
$U=\frac{k q_{1}\, q_{2}}{A B}+\frac{k q_{1} \, q_{3}}{A C}+\frac{k q_{2} \, q_{3}}{B C}$
$=\frac{k\left(4 \times 3 \times 10^{-12}\right)}{4 \times 10^{-2}}+\frac{k\left(4 \times 5 \times 10^{-12}\right)}{5 \times 10^{-2}}+\frac{k\left(3 \times 5 \times 10^{-12}\right)}{3 \times 10^{-2}}$
$=k\left[3 \times 10^{-10}+4 \times 10^{-10}+5 \times 10^{-10}\right]$
$=9 \times 10^{9} \times 12 \times 10^{-10}$
$=108 \times 10^{-1}=10.8\, J$

When three charges located of an equilateral triangle of side $3 \, cm$, the final potential energy of three charges system,
$=\frac{k q_{1}\, q_{2}}{A B}+\frac{k q_{2} \, q_{3}}{B C}+\frac{k q_{1} \, q_{3}}{A C} $
$=\frac{k\left(4 \times 3 \times 10^{-12}\right)}{3 \times 10^{-2}}+\frac{k\left(3 \times 5 \times 10^{-12}\right)}{3 \times 10^{-2}}+\frac{k\left(4 \times 5 \times 10^{-12}\right)}{3 \times 10^{-2}} $
$=\frac{9 \times 10^{9}}{3}\left[12 \times 10^{-10}+15 \times 10^{-10}+20 \times 10^{-10}\right]=14.1 \, J$
Hence, the work done in moving charges at points $A$ and $C, W=(14.1-10.8)=3.3\, J$