Question

Charges +2uc and +3uc lies on the x-y płane at positions (0,0)and (16,0) cm-respectively.

Compute the electric field at point (8, 6) cm.

Answer 1

By the superposition property, the resulting field in $\vec{E}$ is the sum $\vec{E}_1+\vec{E}_2$ of fields generated by each of charges. Let us calculate each of fields :

• $\vec{E}_1=\frac{1}{4\pi \varepsilon_0}\frac{q_1}{|r_1-r|^3}(\vec{r}-\vec{r}_1)=\frac{1}{4\pi \cdot 8.85\cdot 10^{-12}}\frac{2\cdot 10^{-6}}{\sqrt{0.08^2+0.06^2}^3}\cdot(0.08,0.06)$

$\vec{E}_1=\frac{9\cdot 10^9 \cdot 2\cdot 10^{-6}\cdot 10^{-2}}{10^{-3}}\cdot (8,6)=(144, 108) \: V/m$

• $\vec{E}_2=\frac{1}{4\pi \varepsilon_0}\frac{q_2}{|r_2-r|^3}(\vec{r}-\vec{r}_2)=\frac{1}{4\pi \cdot 8.85\cdot 10^{-12}}\frac{3\cdot 10^{-6}}{\sqrt{0.08^2+0.06^2}^3}\cdot(-0.08,0.06)$

$\vec{E}_2=\frac{9\cdot 10^9 \cdot 3\cdot 10^{-6}\cdot 10^{-2}}{10^{-3}}\cdot (-8,6)=(-216, 162) \: V/m$

Therefore, the total field is $\vec{E}=(-72,270)\: V/m$