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2. Electric Potential and Capacitance
hard
For equal point charges $Q$ each are placed in the $xy$ plane at $(0, 2), (4, 2), (4, -2)$ and $(0, -2)$. The work required to put a fifth change $Q$ at the origin of the coordinate system will be
A
$\frac{{{Q^2}}}{{4\pi {\varepsilon _0}}}\left( {1 + \frac{1}{{\sqrt 3 }}} \right)$
B
$\frac{{{Q^2}}}{{4\pi {\varepsilon _0}}}\left( {1 + \frac{1}{{\sqrt 5 }}} \right)$
C
$\frac{{{Q^2}}}{{2\sqrt 2 \pi {\varepsilon _0}}}$
D
$\frac{{{Q^2}}}{{4\pi {\varepsilon _0}}}$
(JEE MAIN-2019)
Solution
$W=V Q=\frac{1}{4 \pi \varepsilon_{0}} Q^{2}\left[\frac{1}{2}+\frac{1}{2}+\frac{2}{2 \sqrt{5}}\right]$
$\therefore \quad \frac{Q^{2}}{4 \pi \varepsilon_{0}}\left[1+\frac{1}{\sqrt{5}}\right]$
Standard 12
Physics
