Answer to Question #104640 in Electric Circuits for Rahul

Question

Consider q1=q2=-q3=2 micro coulomb.APC is an equilateral triangle of side 4cm,B is the middle point of AC.q1, q2 and q3 are situated at the points A, B, C respectively. Determine the magnitude and direction of the electric field at point P.

Answer 1

$\lvert q_1\rvert=\lvert q_2\rvert=\lvert q_3\rvert=q$

$r=4cm$

$\lvert \overrightarrow{E}_1\rvert=\lvert \overrightarrow{E}_3\rvert=E=\frac{1}{4\pi\epsilon_0}\cdot \frac{q}{r^2}=\frac{1}{4\cdot 3.14\cdot 8.85\cdot 10^{-12}}\cdot \frac{2\cdot10^{-6}}{0.04^2}=11.25\cdot 10^6V/m$

$\lvert \overrightarrow{E}_2\rvert=\frac{1}{4\pi\epsilon_0}\cdot \frac{q}{(r\cdot \sin60°)^2}=\frac{1}{4\cdot 3.14\cdot 8.85\cdot 10^{-12}}\cdot \frac{2\cdot10^{-6}}{(0.04\cdot \sin60°)^2}=15\cdot 10^6V/m$

$\overrightarrow{E}_{123}=\overrightarrow{E}_1+\overrightarrow{E}_2+\overrightarrow{E}_3$

$\overrightarrow{E}_1+\overrightarrow{E}_3=\overrightarrow{E}_{13}$

$\lvert \overrightarrow{E}_{13}\rvert=2E\cos60°=E=11.25\cdot 10^6V/m$

$\lvert\overrightarrow{E}_{123}\rvert=\sqrt{(11.25\cdot 10^6)^2+(15\cdot 10^6)^2}=18.75\cdot 10^6V/m$ Answer

$\tan\alpha=\frac{11.25}{15}=0.75\to \alpha=36.87°$ Answer