Identical +1.39 µC charges are fixed to adjacent corners of a square. What charge (magnitude and algebraic sign) should be fixed to one of the empty corners, so that the total electric potential at the remaining empty corner is 0 V?

Answer 1

{V_T} &= {V_1} + {V_2} + {V_3}\\
{V_T} &= \dfrac{{k{Q_1}}}{l} + \dfrac{{k{Q_2}}}{{l\sqrt 2 }} + \dfrac{{k{Q_3}}}{l}
0 &= \dfrac{{k{Q_1}}}{l} + \dfrac{{k{Q_2}}}{{l\sqrt 2 }} + \dfrac{{k{Q_3}}}{l}\\
\dfrac{k}{l}\left( {{Q_1} + \dfrac{{{Q_2}}}{{\sqrt 2 }} + {Q_3}} \right) = 0\\
{Q_3} + {Q_1} + \dfrac{{{Q_2}}}{{\sqrt 2 }} &= 0\\
{Q_3}& = - \left( {7.74 \times {{10}^{ - 6}}\;{\rm{C + }}\dfrac{{7.74 \times {{10}^{ - 6}}\;{\rm{C}}}}{{\sqrt 2 }}} \right)\\
{Q_3}& = - 1.32 \times {10^{ - 5}}\;{\rm{C}}