Question 20.1: Four point charges q1 = q2 = Q and q3 = q4 = −Q, where Q = √......
Four point charges q_{1} = q_{2} = Q and q_{3} = q_{4} = −Q, where Q = \sqrt{2} μC, are placed at the four corners of a square of side a = 0.4 m, see Fig. 20.5a. Find the electric field at the center P of the square.
The distance between each charge and the center P of the square is a/\sqrt{2}. At point P, the point charges q_{1} and q_{3} produce two diagonal electric field vectors \vec{E}_{1} and \vec{E}_{3}, both directed toward q_{3}, see Fig. 20.5b. Hence, their vector sum \vec{E}_{13} = \vec{E}_{1} + \vec{E}_{3} points toward q_{3} and has the magnitude:
E_{13}=E_{1}+E_{3}=k\frac{Q}{(a/\sqrt{2})^{2}}+k\frac{Q}{(a/\sqrt{2})^{2}}=4k\frac{Q}{a^{2}}At point P, the charges q_{2} and q_{4} produce two diagonal electric fields \vec{E}_{2} and \vec{E}_{4}, both directed toward q_{4}, see Fig. 20.5b. Hence, their vector sum \vec{E}_{24} = \vec{E}_{2} + \vec{E}_{4} points toward q_{4} and has the magnitude:
E_{24}=E_{2}+E_{4}=k{\frac{Q}{(a/{\sqrt{2}})^{2}}}+k{\frac{Q}{(a/{\sqrt{2}})^{2}}}=4k{\frac{Q}{a^{2}}}We now must combine the two electric field vectors \vec{E}_{13} and \vec{E}_{24} to form the resultant electric field vector \vec{E} = \vec{E}_{13} + \vec{E}_{24} which is along the positive x-direction and has the magnitude:
E=E_{13}\;\mathrm{cos}\,45°+E_{24}\;\mathrm{cos}\,45°=2\times{\Biggl(}4k{\frac{Q}{a^{2}}}\times{\frac{1}{\sqrt{2}}}{\Biggl)}=k{\frac{8Q}{\sqrt{2}a^{2}}}=(9\times10^{9}\,\mathrm{N.m^{2}}/\mathrm{C^{2}})\frac{8(\sqrt{2}\times10^{-6}\,{\mathrm{m}})}{\sqrt{2}\;(0.4\,{\mathrm{m}})^{2}}=4.5\times10^{5}\,{\mathrm{N/C}}
