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?

Question

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?

Accepted Answer

point, charge location [latex]+Q A\left(\frac{-9}{2}, 0\right),+Q B\left(\frac{9}{5}, 0\right)[/latex] and [latex]-2 Q C\left(0, \frac{\sqrt{3}}{2} a\right)[/latex]

(i) Let, P(a, y) be the point on y-axis, where [latex]V=V_{1}+V_{2}+V_{3}[/latex] due to these change is zero, [latex]V=V_{1}+V_{2}+V_{3}=0[/latex]

[latex]\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}[/latex]

(ii) [latex]\underline{V}[/latex] at point P(0, y)

[latex]\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}[/latex]

[latex]\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}[/latex]

Solving

[latex]\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][/latex]

[latex]\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][/latex]

[latex]\text { At } P(a, y)=P\left(o, \frac{a}{2 \sqrt{3}}\right)[/latex]

[latex]\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}[/latex]

[latex]E=E_{y}=\frac{3 Q}{4 \pi E_{0} a^{2}}[/latex]

(iii) The direction of [latex]\vec{E}[/latex] 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 [latex]\vec{E}[/latex] is in the [latex]-\nu e y[/latex] direction, the angle between the equipotential surface and y-axis is zero.

Question 1.5.3: Three electrostatic point charges are located in the xyplane...

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