If \(A B+A C=12 \mathrm{~cm}\) and \(A B \cdot A C=32 \mathrm{~cm}^{2}\), the potential energy of the charge at \(A\) in Joules is (round off to the nearest integer)
Three identical charge each of \(2 \mu \mathrm{C}\) are placed at the vertices of a triangle…
If \(A B+A C=12 \mathrm{~cm}\) and \(A B \cdot A C=32 \mathrm{~cm}^{2}\), the potential energy of the charge at \(A\) in Joules is (round off to the nearest integer)
Solution
\(A B \cdot A C=32 \mathrm{~cm}^{2}\)
\(\begin{array}{l}
\therefore \quad A B-A C=\sqrt{(A B+A C)^{2}-4 A B \cdot A C} \\
A B-A C=4 \quad \dots\text{(ii)}
\end{array}\)
From equation (i) and (ii)
\(A B=8 \mathrm{~cm} ; A C=4 \mathrm{~cm}\)
Potential energy at point \(A\)
\(\begin{array}{l}
V_{A}=\frac{1}{4 \pi \varepsilon_{0}} q_{1} q_{2}\left[\frac{1}{A B}+\frac{1}{A C}\right] \\
=\frac{9 \times 10^{9} \times 2 \times 2 \times 10^{-12}}{10^{-2}}\left(\frac{1}{8}+\frac{1}{4}\right)=1.35 J
\end{array}\)
Asked in: JEE Mains - Electrostatics - Chapter Test