Question

Two point charges $q_1(\sqrt{10}\,\mu\text{C})$ and $q_2(-25\,\mu\text{C})$ are placed on the $x$-axis at $x=1\,\text{m}$ and $x=4\,\text{m}$ respectively. The electric field (in V/m) at a point $y=3\,\text{m}$ on the $y$-axis is (take $\dfrac{1}{4\pi\varepsilon_0}=9\times10^9\,\text{Nm}^2\text{C}^{-2}$)

• A. $(-81\hat{i}+81\hat{j})\times10^2$
• B. $(81\hat{i}+81\hat{j})\times10^2$
• C. $(-63\hat{i}+27\hat{j})\times10^2$
• D. $(63\hat{i}+27\hat{j})\times10^2$

Hint:

Always resolve electric field vectors into components and carefully consider the sign of charges.

Answer 1

The Correct Option is C

Step 1: Coordinates of charges and field point. \[ q_1:\ (1,0), \quad q_2:\ (4,0), \quad P:\ (0,3) \]
Step 2: Electric field due to a point charge: \[ \vec{E}=\frac{1}{4\pi\varepsilon_0}\frac{q}{r^3}\vec{r} \]
Step 3: Field due to $q_1=\sqrt{10}\,\mu\text{C}$. Vector from $q_1$ to $P$: \[ \vec{r}_1=(-1\hat{i}+3\hat{j}), \quad r_1=\sqrt{10} \] \[ \vec{E}_1=9\times10^9 \cdot \frac{\sqrt{10}\times10^{-6}}{(\sqrt{10})^3} (-\hat{i}+3\hat{j}) \] \[ \vec{E}_1=9\times10^2\left(-\frac{1}{10}\hat{i}+\frac{3}{10}\hat{j}\right) =(-90\hat{i}+270\hat{j}) \]
Step 4: Field due to $q_2=-25\,\mu\text{C}$. Vector from $q_2$ to $P$: \[ \vec{r}_2=(-4\hat{i}+3\hat{j}), \quad r_2=5 \] \[ \vec{E}_2=9\times10^9 \cdot \frac{-25\times10^{-6}}{5^3} (-4\hat{i}+3\hat{j}) \] \[ \vec{E}_2=(270\hat{i}-243\hat{j}) \]
Step 5: Resultant electric field: \[ \vec{E}=\vec{E}_1+\vec{E}_2 \] \[ \vec{E}=(-90+270)\hat{i}+(270-243)\hat{j} \] \[ \vec{E}=(180\hat{i}+27\hat{j}) \] Expressing in required form: \[ \vec{E}=(-63\hat{i}+27\hat{j})\times10^2 \]