Electrostatics 1 Question 3
3. Two point charges $q _1(\sqrt{10} \mu C)$ and $q _2(-25 \mu C)$ are placed on the $x$-axis at $x=1 m$ and $x=4 m$, respectively. The electric field (in V/m) at a point $y=3 m$ on $Y$-axis is
(Main 2019, 9 Jan II)
Take, $\frac{1}{4 \pi \varepsilon _0}=9 \times 10^{9} N-m^{2} C^{-2}$
(a) $(63 \hat{\mathbf{i}}-27 \hat{\mathbf{j}}) \times 10^{2}$
(b) $(81 \hat{\mathbf{i}}-81 \hat{\mathbf{j}}) \times 10^{2}$
(c) $(-81 \hat{\mathbf{i}}+81 \hat{\mathbf{j}}) \times 10^{2}$
(d) $(-63 \hat{\mathbf{i}}+27 \hat{\mathbf{j}}) \times 10^{2}$
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Answer:
Correct Answer: 3. (d)
Solution:
- Here, $q _1=\sqrt{10} \mu C=\sqrt{10} \times 10^{-6} C$
$q _2=-25 \mu C=-25 \times 10^{-6} C$
Let $\mathbf{E} _1$ and $\mathbf{E} _2$ are the values of electric field due to $q _1$ and $q _2$ respectively.
$$ \begin{aligned} \text { Here, } E _1 & =\frac{1}{4 \pi \varepsilon _0} \cdot \frac{q _1}{A C^{2}}=\frac{1}{4 \pi \varepsilon _0} \times \frac{\sqrt{10} \times 10^{-6}}{\left(1^{2}+3^{2}\right)} \\ & =9 \times 10^{9} \times \sqrt{10} \times 10^{-7} \\ & =9 \sqrt{10} \times 10^{2} \\ \therefore \quad \mathbf{E} _1 & =9 \sqrt{10} \times 10^{2}\left[\cos \theta _1(\hat{\mathbf{i}})+\sin \theta _1 \hat{\mathbf{j}}\right] \end{aligned} $$
From $\triangle O A C$,
$$ \begin{aligned} & \sin \theta _1=\frac{3}{\sqrt{10}} \text { and } \cos \theta _1=\frac{1}{\sqrt{10}} \\ & \therefore E _1=9 \sqrt{10} \times 10^{2} \frac{1}{\sqrt{10}}(\hat{\mathbf{i}})+\frac{3}{\sqrt{10}} \hat{\mathbf{j}} \\ &=9 \times 10^{2}[\hat{\mathbf{i}}+3 \hat{\mathbf{j}}] \\ &=(-\hat{\mathbf{i}}+27 \hat{\mathbf{j}}) \times 10^{2} V / m \end{aligned} $$
and $E _2=\frac{1}{4 \pi \varepsilon _0} \cdot \frac{-25 \times 10^{-6}}{\left(4^{2}+3^{2}\right)}=9 \times 10^{3} V / m$
From $\triangle O B C$,
$$ \begin{aligned} & \sin \theta _2=\frac{3}{5} \\ & \cos \theta _2=\frac{4}{5} \\ \therefore \quad & \mathbf{E} _2=9 \times 10^{3}\left[\cos \theta _2 \hat{\mathbf{i}}-\sin \theta _2 \hat{\mathbf{j}}\right] \\ & \mathbf{E} _2=9 \times 10^{3} \frac{4}{5} \hat{\mathbf{i}}-\frac{3}{5} \hat{\mathbf{j}}=(72 \hat{\mathbf{i}}-54 \hat{\mathbf{j}}) \times 10^{2} \\ \therefore \quad & \mathbf{E}=\mathbf{E} _1+\mathbf{E} _2=(63 \hat{\mathbf{i}}-27 \hat{\mathbf{j}}) \times 10^{2} V / m \end{aligned} $$
