Question

For the electric field shown in figure, its vector representation will be

• A. $\overrightarrow{E}=200\hat{i}-\frac{200}{\sqrt{3}}\hat{j}$
• B. $\overrightarrow{E}=100\hat{i}-100\sqrt{3}\hat{j}$
• C. $\overrightarrow{E}=\frac{50}{\sqrt{3}}\hat{i}-\frac{100}{\sqrt{3}}\hat{j}$
• D. $\overrightarrow{E}=10\hat{i}-5\sqrt{2}\hat{j}$

Answer 1

The correct option is A $\overrightarrow{E}=200\hat{i}-\frac{200}{\sqrt{3}}\hat{j}$

From the given diagram, the electric filed line are parallel and equally spaced. It means $\overrightarrow{E}$ is uniform.
The planes perpendicular to the direction of electric field represents an equipotential surface of $40\text{V},20\text{V},\&0\text{V}$


Potential difference between $\text{A}$ and $\text{B}$

$\Delta V=V_A-V_B=20V$

$\Rightarrow$ Distance between equipotential surfaces along direction of $\overrightarrow{E}$,

$d=\left(10\text{cm}\right)\cos {30}^{\circ }$

$\Rightarrow d=0.1\times \frac{\sqrt{3}}{2}$

Therefore, we know that

$V_A-V_B=Ed$

$\Rightarrow 20=E\times 0.1\times \frac{\sqrt{3}}{2}$

$\Rightarrow E=\frac{400}{\sqrt{3}}$

Now, component of electric field

$E_x=E\cos {30}^{\circ }\hat{i}$

$\Rightarrow E_x=(\frac{400}{\sqrt{3}}\times \frac{\sqrt{3}}{2})\hat{i}=200\hat{i}$

$E_y=-E\sin {30}^{\circ }\hat{j}=-\frac{200}{\sqrt{3}}\hat{j}$

$\therefore \overrightarrow{E}=200\hat{i}-\frac{200}{\sqrt{3}}\hat{j}$

Therefore, option $\left(a\right)$ is correct.

Tip : Uniform electric field is characterized by equally spaced parallel field lines. For uniform electric field, we can apply $\Delta V=Ed$, where $d$ is the distance between two points along $\overrightarrow{E}$.