Question

A +11 nC charge is located at (x,y)=(0cm,12cm) and a -3.0 nC charge is located at (x,y)=(7.0cm,0cm).

Where would a -14 nC charge need to be located in order that the electric field at the origin be zero?

Answer 1

Electric field ,$E=kQ/r^{2}$

$E_{1}=k*11nC/(0.12m)^{2}$

$E_{1}=9*10^{9}*1.1*10^{-8}/0.12^{2}$

$E_{1}=6875N/C$

${\color{Red} \vec{E_{1}}=-6875\hat{j}\ N/C}$

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$E_{2}=k*3nC/(0.07m)^{2}$

$E_{2}=9*10^{9}*3*10^{-9}/0.07^{2}$

$E_{2}=5510.2N/C$

${\color{Red} \vec{E_{2}}=5510.2\hat{i}\ N/C}$

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$\vec{E_{3}}=E_{3x}\hat{i}+E_{3y}\hat{j}$

Given $\vec{E_{1}}+\vec{E_{2}}+\vec{E_{3}}=0$

$-6875\hat{j}+5510.2\hat{i}+E_{3x}\hat{i}+E_{3y}\hat{j}=0$

$(5510.2+E_{3x})\hat{i}+(-6875+E_{3y})\hat{j}=0$

$(5510.2+E_{3x})=0$

$E_{3x}=-5510.2$

$kq_{3}/x^{2}=-5510.2$

$9*10^{9}*1.4*10^{-8}/x^{2}=-5510.2$

$9*10^{9}*1.4*10^{-8}/5510.2=x^{2}$

Ignore negative sign .it just shows x is negative

$x=0.1512m=15.12cm$

ANSWER: ${\color{Red} x=-15.12cm}$

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Similarly

$(5510.2+E_{3x})\hat{i}+(-6875+E_{3y})\hat{j}=0$

$-6875+E_{3y}=0$

$6875=E_{3y}$

$6875=kq_{3}/y^{2}$

$6875=9*10^{9}*1.4*10^{-8}/y^{2}$

$y^{2}=9*10^{9}*1.4*10^{-8}/6875$

$y=0.1354m=13.54cm$

ANSWER: ${\color{Red} y=13.54cm}$

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ANSWER: ${\color{Red} (-15.12cm,13.54cm)}$

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