JEE Mains · Physics · STD 12 - 2. Electric potential and capacitance
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 \(\frac{{{Q^2}}}{{4\pi {\varepsilon _0}}}\)
Answer & Solution
Correct Answer
(B) \(\frac{{{Q^2}}}{{4\pi {\varepsilon _0}}}\left( {1 + \frac{1}{{\sqrt 5 }}} \right)\)
Step-by-step Solution
Detailed explanation
\(W=V Q=\frac{1}{4 \pi \varepsilon_{0}} Q^{2}\left[\frac{1}{2}+\frac{1}{2}+\frac{2}{2 \sqrt{5}}\right]\) \(\therefore \quad \frac{Q^{2}}{4 \pi \varepsilon_{0}}\left[1+\frac{1}{\sqrt{5}}\right]\)
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