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Four electric charges $+q,+q, -q$ and $-q$ are placed at the comers of a square of side $2L$ (see figure). The electric potential at point $A,$ midway between the two charges $+q$ and $+q,$ is
$\;\frac{1}{{4\pi {\varepsilon _0}}}\frac{{2q}}{L}\left( {1 + \sqrt 5 } \right)$
$\;\frac{1}{{4\pi {\varepsilon _0}}}\frac{{2q}}{L}\left( {1 + \frac{1}{{\sqrt 5 }}} \right)$
$\;\frac{1}{{4\pi {\varepsilon _0}}}\frac{{2q}}{L}\left( {1 - \frac{1}{{\sqrt 5 }}} \right)$
zero
Solution
$A$ is the midpoint
of $PS$
${\therefore \quad PA = AS = L}$
${AR = AQ = \sqrt {{{(SR)}^2} + {{(AS)}^2}} }$
${ = \sqrt {{{(2L)}^2} + {{(L)}^2}} = L\sqrt 5 }$
Electric potential at point $A$ due to the given charge configuration is
$V_{A} =\frac{1}{4 \pi \varepsilon_{0}}\left[\frac{q}{P A}+\frac{q}{A S}+\frac{(-q)}{A Q}+\frac{(-q)}{A R}\right]$
$=\frac{1}{4 \pi \varepsilon_{0}}\left[\frac{q}{L}+\frac{q}{L}-\frac{q}{L \sqrt{5}}-\frac{q}{L \sqrt{5}}\right]$
$=\frac{1}{4 \pi \varepsilon_{0}}\left[\frac{2 q}{L}-\frac{2 q}{L \sqrt{5}}\right]=\frac{1}{4 \pi \varepsilon_{0}} \frac{2 q}{L}\left[1-\frac{1}{\sqrt{5}}\right]$
