Three electrostatic point charges are located in the xy-plane as given below: +Q at (-a/2, 0), + Q at (a/2, 0) and – 02Q at (0, > /3/2)
Calculate the coordinate of the point, P, on the y-axis, where the potential due to these charges is zero. Also, calculate the magnitude of the electric field strength at P. At the point, P what is the angle between the equipotential passing through P and the y-axis?
point, charge location +Q A\left(\frac{-9}{2}, 0\right),+Q B\left(\frac{9}{5}, 0\right) and -2 Q C\left(0, \frac{\sqrt{3}}{2} a\right)
(i) Let, P(a, y) be the point on y-axis, where V=V_{1}+V_{2}+V_{3} due to these change is zero, V=V_{1}+V_{2}+V_{3}=0
\begin{aligned}&\frac{Q}{4 \pi E_{0}}\left[\frac{1}{A P}+\frac{1}{B P}-\frac{2}{C P}\right]=0 \\&\text { But } A B=B P, \text { so, } \\&\frac{1}{A P}+\frac{1}{A P}-\frac{2}{C P} \text { or } A P=C P \\&\sqrt{\frac{a^{2}}{4}+y^{2}}=y-\frac{\sqrt{3}}{2} a, \Rightarrow y=\frac{a}{2 \sqrt{3}} \\&P(a, y)=P\left(0, \frac{a}{2 \sqrt{3}}\right)\end{aligned}
(ii) \underline{V} at point P(0, y)
\begin{aligned}V &=\frac{Q}{4 \pi E_{0}}\left[\frac{1}{A P}+\frac{1}{B P}-\frac{2}{C P}\right] \\&=\frac{Q}{2 \pi E_{0}}\left[\frac{1}{A P}-\frac{1}{C P}\right]\end{aligned}\begin{aligned}&V=\frac{Q}{2 \pi E_{0}}\left[\frac{1}{\sqrt{\frac{a^{2}}{4}+y^{2}}}-\frac{1}{\sqrt{y-\frac{\sqrt{3}}{2}} a}\right] \\&\vec{E}=-\nabla V=-\left[\frac{\partial}{\partial x} V \hat{a}_{x}+\frac{\partial}{\partial y} V \hat{a}_{y}+\frac{\partial}{\partial z} V \hat{a}_{z}\right] \\&\frac{\partial V}{\partial x}=0 \\&\frac{\partial V}{\partial z}=0\end{aligned}
Solving
\frac{\partial \nu }{\partial y}=\frac{Q}{2 \pi E_{0}}\left[\begin{array}{l}-\frac{1}{2}\left(\frac{a^{2}}{4}+y^{2}\right)^{-3 / 2} \\(2 y)+\left(y-\frac{\sqrt{3}}{2} a\right)^{-2}\end{array}\right]\vec{E}=\frac{-Q}{2 \pi E_{0}}\left[\begin{array}{l}-y\left(\frac{a^{2}}{4}+y^{2}\right)^{-3 / 2} \\+\left(y-\frac{\sqrt{3}}{2} a\right)^{-2} \vec{a}_{y}\end{array}\right]
\text { At } P(a, y)=P\left(o, \frac{a}{2 \sqrt{3}}\right)
\vec{E}=\frac{-Q}{2 \pi E_{0}}\left[\begin{array}{l}\frac{-a}{2 \sqrt{3}}\left(\frac{a^{2}}{4}+\frac{a^{2}}{12}\right)^{-3 / 2} \\+\left(\frac{a}{2 \sqrt{3}}-\frac{\sqrt{3}}{2} a\right)^{-2}\end{array}\right] \vec{a}_{x}
E=E_{y}=\frac{3 Q}{4 \pi E_{0} a^{2}}
(iii) The direction of \vec{E} at P is the direction of the normal to the equipotential surface (V = 0) at that points in the direction of the decreasing values of V. The direction of \vec{E} is in the -\nu e y direction, the angle between the equipotential surface and y-axis is zero.
