Three point charges of 0.33×10^(-8), 0.33×10^(-8) and 0.65×10^(-8) are at the points A,B,C, of a square ABCD. Find the electric field at the corner D?

Dear student,

Electric field at point D will the sum of the electric fields due to A,B and C

Electric field at D if l is the length of the side of the square$$\begin{gathered} E=k\left[\frac{q_1}{l^2}+\frac{q_2}{2l^2}+\frac{q_3}{l^2}\right] \\ The charge at B will be diagonal \\ and so the dis\tan ce will be 2l^2 \\ Taking the components of the electric field along the diagonal \\ E= k\left[\frac{0.33\times {10}^{-8}}{l^2}\cos 45+\frac{0.33\times {10}^{-8}}{2l^2}+\frac{0.65\times {10}^{-8}}{l^2}\cos 45\right] \\ E=k\left[\frac{0.33\times {10}^{-8}\times 0.707}{l^2}+\frac{0.33\times {10}^{-8}}{2l^2}+\frac{0.65\times {10}^{-8}\times 0.707}{l^2}\right] \\ E=\frac{9\times {10}^9\times {10}^{-8}}{l^2}\left[0.23+0.165+0.46\right] \\ E=\frac{76.9}{l^2}N/C \end{gathered}$$
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