Question

two point charges q1 and q2 are located at points (a, 0,0)and (0,b,0) respectively.find the electric field due to both these charges, at point (0,0,c).

Answer 1

The electric field caused by the two charges is $\bold{\frac { 1 }{ 4\pi \varepsilon _{ 0 } } \left[ \frac { q_{ 1 }(-a\hat {i} -c\hat { k } ) }{ \left( a^{ 2 }+c^{ 2 } \right) ^{ \frac { 3 }{ 2 } } } +\frac { q_{ 2 }(-b\hat { j } +c\hat { k } ) }{ \left( b^{ 2 }+c^{2}\right) ^{\frac {3}{2}}} \right]}$

Solution:

The electric field caused by these charges will have net field at point (0, 0, c)

$\begin{matrix} E_{net} & =\overrightarrow {E_{1}} +\overrightarrow {E_{2}} \\\\ E_{net} & =\frac {1}{4\pi \varepsilon _{0}} \frac {q_{1}}{r_{1}^{3}} \overrightarrow {AC} +\frac{1}{4\pi \varepsilon _{0}} \frac {q_{2}}{r_{2}^{3}} \overrightarrow {BC} \\ \\\overrightarrow {AC} & =-a\hat{\imath}-c\hat{k} \\ \\AC & =\sqrt{a^{2}+c^{2}} \\ \\\overrightarrow {BC} & =-b\hat {\jmath} +c\hat{k} \\ \\BC & =\sqrt {b^{2}+c^{2}}\end{matrix}$

Thereby, the net field is given by $\vec{E}=\frac{1}{4 \pi \varepsilon_{0}} \frac{q_{1}}{r_{1}^{3}} \overrightarrow{A C}+\frac{1}{4 \pi \varepsilon_{0}} \frac{q_{1}}{r_{2}^{3}} \overrightarrow{B C}=\frac{1}{4 \pi \varepsilon_{0}}\left[\frac{q_{1}(-a \hat{\imath}-c \hat{k})}{\left(a^{2}+c^{2}\right)^{\frac{3}{2}}}+\frac{q_{2}(-b\hat{j}+c \hat{k} )}{\left(b^{2}+c^{2}\right)^{\frac{3}{2}}}\right]$