Question

Find the electric field vector at $\text{P(a, a, a)}$ due to three infinitely long lines of charges along the $\text{x,y}$ and $\text{z}-$ axis, respectively. The charge density, i.e., charge per unit length of each wire is $\lambda$.

• A. $\frac{\lambda }{3\pi {ϵ}_{0}a}(\hat{i}+\hat{j}+\hat{k})$
• B. $\frac{\lambda }{2\pi {ϵ}_{0}a}(\hat{i}+\hat{j}+\hat{k})$
• C. $\frac{\lambda }{2\sqrt{2}\pi {ϵ}_{0}a}(\hat{i}+\hat{j}+\hat{k})$
• D. $\frac{\sqrt{2}\lambda }{\pi {ϵ}_{0}a}(\hat{i}+\hat{j}+\hat{k})$

Answer 1

The correct option is B $\frac{\lambda }{2\pi {ϵ}_{0}a}(\hat{i}+\hat{j}+\hat{k})$


Let us consider the electric field due to wire $\left(3\right)$ only,

${\overrightarrow{E}}_3=E\hat{u}$

where,
$\hat{u}=(\hat{i}\cos {45}^{\circ }+\hat{j}\cos {45}^{\circ })$ is the unit vector,

$\overrightarrow{E}=\frac{\lambda }{2\pi {ϵ}_{0}OP}=\frac{\lambda }{2\pi {ϵ}_0(a^2+a^2{)}^{\frac{1}{2}}}$

Substituting the value of $E$ and $\hat{u}$, we get
${\overrightarrow{E}}_3=\frac{\lambda }{2\pi {ϵ}_0(a^2+a^2{)}^{\frac{1}{2}}}(\hat{i}\cos {45}^{\circ }+\hat{j}\cos {45}^{\circ })$

$\Rightarrow {\overrightarrow{E}}_3=\frac{\lambda }{2\pi {ϵ}_{0}a\sqrt{2}}\times \frac{1}{\sqrt{2}}(\hat{i}+\hat{j})$

$\Rightarrow {\overrightarrow{E}}_3=\frac{\lambda }{4\pi {ϵ}_{0}a}(\hat{i}+\hat{j})$

Similarly, electric field due to wires $\left(1\right)$ and $\left(2\right)$,

${\overrightarrow{E}}_1=\frac{\lambda }{4\pi {ϵ}_{0}a}(\hat{j}+\hat{k})$

${\overrightarrow{E}}_2=\frac{\lambda }{4\pi {ϵ}_{0}a}(\hat{i}+\hat{k})$

Thus, net electric field at $\text{P}$,

${\overrightarrow{E}}_{\text{net}}={\overrightarrow{E}}_1+{\overrightarrow{E}}_2+{\overrightarrow{E}}_3$

$\therefore {\overrightarrow{E}}_{\text{net}}=\frac{\lambda }{2\pi {ϵ}_{0}a}(\hat{i}+\hat{j}+\hat{k})$

Hence, option (b) is the correct answer.