Two point charges 3uC and 2.5uC are places at point A(1, 1,2) m and B(0, 3, –1) m respectively. Find out electric field intensity at point C(3,3,3) m.

Dear Student,

Please find below the solution to the asked query:

Given that the two charges $3 \mu C \& 2.5 \mu C$ are at $A\left(1,1,2\right) \& B\left(0,3,-1\right)$ respectively.

The electric field intensity at the point $C\left(3,3,3\right)$

Due to Charge at A is,

$$\begin{gathered} {\overrightarrow{E}}_1=\frac{1}{4\pi {\epsilon }_0}\frac{3\times {10}^{-6}}{r^3}\overrightarrow{r} \\ \Rightarrow {\overrightarrow{E}}_1=\frac{1}{4\pi {\epsilon }_0}\frac{3\times {10}^{-6}}{{\left(\sqrt{{\left(3-1\right)}^2+{\left(3-1\right)}^2+{\left(3-2\right)}^2}\right)}^3}\left(\left(3-1\right) \hat{i} + \left(3-1\right) \hat{j} + \left(3-2\right) \hat{k}\right) \\ \Rightarrow {\overrightarrow{E}}_1=\frac{9\times {10}^9\times 3\times {10}^{-6}}{{\left(\sqrt{4+4+1}\right)}^3}\left(2 \hat{i} + 2 \hat{j} + \hat{k}\right) \\ \Rightarrow {\overrightarrow{E}}_1=\frac{27\times {10}^3}{27}\left(2 \hat{i} + 2 \hat{j} + \hat{k}\right) \\ \Rightarrow {\overrightarrow{E}}_1=\left(2 \hat{i} + 2 \hat{j} + \hat{k}\right)\times {10}^3 \raisebox{1ex}{$N$}\!\left/ \!\raisebox{-1ex}{$C$}\right. \end{gathered}$$

Due to charge at B is,

$$\begin{gathered} {\overrightarrow{E}}_2=\frac{1}{4\pi {\epsilon }_0}\frac{2.5\times {10}^{-6}}{r^3}\overrightarrow{r} \\ \Rightarrow {\overrightarrow{E}}_2=\frac{1}{4\pi {\epsilon }_0}\frac{2.5\times {10}^{-6}}{{\left(\sqrt{{\left(3-0\right)}^2+{\left(3-3\right)}^2+{\left(3+1\right)}^2}\right)}^3}\left(\left(3-0\right) \hat{i} + \left(3-3\right) \hat{j} + \left(3+1\right) \hat{k}\right) \\ \Rightarrow {\overrightarrow{E}}_2=\frac{9\times {10}^9\times 2.5\times {10}^{-6}}{{\left(\sqrt{9+16}\right)}^3}\left(3 \hat{i} +4 \hat{k}\right) \\ \Rightarrow {\overrightarrow{E}}_2=\frac{22.5\times {10}^3}{125}\left(3 \hat{i} +4 \hat{k}\right) \\ \Rightarrow {\overrightarrow{E}}_2=\left(3 \hat{i} +4 \hat{k}\right)\times 0.18\times {10}^3 \raisebox{1ex}{$N$}\!\left/ \!\raisebox{-1ex}{$C$}\right. \end{gathered}$$

Therefore, according to the superposition principle, the net electric field intensity at C is,

$$\begin{gathered} \overrightarrow{E}={\overrightarrow{E}}_1+{\overrightarrow{E}}_2=\left(2 \hat{i} + 2 \hat{j} + \hat{k}\right)\times {10}^3+\left(3 \hat{i} + \hat{k}\right)\times 0.18\times {10}^3 \\ \Rightarrow \overrightarrow{E}=\left[\left(2+0.54\right) \hat{i} + \left(2+0\right) \hat{j} +\left(1+0.18\right) \hat{k}\right] \times {10}^3 \\ \Rightarrow \overrightarrow{E}=\left[2.54 \hat{i} +2 \hat{j} +1.18 \hat{k}\right] \raisebox{1ex}{$kN$}\!\left/ \!\raisebox{-1ex}{$C$}\right. \\ \Rightarrow \left|\overrightarrow{E}\right|=\sqrt{{\left(2.54\right)}^2+{\left(2\right)}^2+{\left(1.18\right)}^2} \raisebox{1ex}{$kN$}\!\left/ \!\raisebox{-1ex}{$C$}\right. \\ \Rightarrow \left|\overrightarrow{E}\right|=\sqrt{6.45+4+1.39} \raisebox{1ex}{$kN$}\!\left/ \!\raisebox{-1ex}{$C$}\right. \\ \Rightarrow \left|\overrightarrow{E}\right|=\sqrt{11.844} \raisebox{1ex}{$kN$}\!\left/ \!\raisebox{-1ex}{$C$}\right. \\ \Rightarrow \left|\overrightarrow{E}\right|=3.44 \raisebox{1ex}{$kN$}\!\left/ \!\raisebox{-1ex}{$C$}\right. \end{gathered}$$

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