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 $ABC$ 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 $3cm$ 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 $\Delta ABC$
$A{C}^2=A{B}^2+B{C}^2$
$\Rightarrow AC=\sqrt{A{B}^2+B{C}^2}=\sqrt{{\left(4\times 10^{-2}\right)}^2+{\left(3\times 10^{-2}\right)}^2}$
$\Rightarrow AC=5\times 10^{-2}m$
Initial electric potential energy of three charges,
$U=\frac{k{q}_1{q}_2}{AB}+\frac{k{q}_1{q}_3}{AC}+\frac{k{q}_2{q}_3}{BC}$
$=\frac{k\left(4\times 3\times 10^{-12}\right)}{4\times 10^{-2}}+\frac{k\left(4\times 5\times 10^{-12}\right)}{5\times 10^{-2}}+\frac{k\left(3\times 5\times 10^{-12}\right)}{3\times 10^{-2}}$
$=k\left[3\times 10^{-10}+4\times 10^{-10}+5\times 10^{-10}\right]$
$=9\times 10^9\times 12\times 10^{-10}$
$=108\times 10^{-1}=10.8J$

When three charges located of an equilateral triangle of side $3cm$, the final potential energy of three charges system,
$=\frac{k{q}_1{q}_2}{AB}+\frac{k{q}_2{q}_3}{BC}+\frac{k{q}_1{q}_3}{AC}$
$=\frac{k\left(4\times 3\times 10^{-12}\right)}{3\times 10^{-2}}+\frac{k\left(3\times 5\times 10^{-12}\right)}{3\times 10^{-2}}+\frac{k\left(4\times 5\times 10^{-12}\right)}{3\times 10^{-2}}$
$=\frac{9\times 10^9}{3}\left[12\times 10^{-10}+15\times 10^{-10}+20\times 10^{-10}\right]=14.1J$
Hence, the work done in moving charges at points $A$ and $C,W=(14.1-10.8)=3.3J$