A uniform electric field of $500 \mathrm{Vm}^{-1}$ is directed at $30^{\circ}$ with the positive $X$-axis as…
- $-250(3 \sqrt{3}+5) \vee$
- $250(3 \sqrt{3}+5) \vee$
- $-250(3+5 \sqrt{3}) V$
- $250(3+5 \sqrt{3}) V$
Solution
$\begin{aligned} \mathbf{E} & =\left(E_0 \cos \theta \hat{\mathbf{i}}+E_0 \sin \theta \hat{\mathbf{j}}\right) \mathrm{Vm}^{-1} \\ O A & =3 \mathrm{~cm} \text { and } O B=5 \mathrm{~m}\end{aligned}$
Now, displacement vector,
$\mathbf{A B}=d \mathbf{r}=(3 \hat{\mathbf{i}}+5 \hat{\mathbf{j}}) \mathrm{m}$
We know that, potential difference,
$\Delta V=-\mathbf{E} \cdot d \mathbf{r}$
i.e. $V_B-V_A=\left(E_0 \cos \theta \hat{\mathbf{i}}+E_0 \sin \theta \hat{\mathbf{j}}\right) \cdot(3 \hat{\mathbf{i}}+5 \hat{\mathbf{j}})$
Substituting the values, we get
$V_B-V_A=-500\left(\cos 30^{\circ} \hat{\mathbf{i}}+\sin 30^{\circ} \hat{\mathbf{j}}\right) \cdot(3 \hat{\mathbf{i}}+5 \hat{\mathbf{j}})$
$\begin{aligned} & =-500\left(3 \cos 30^{\circ}+5 \sin 30^{\circ}\right) \\ & =-500\left[\frac{3 \sqrt{3}}{2}+\frac{5}{2}\right] \mathrm{V}\end{aligned}$
$\therefore \quad V_B-V_A=-250(3 \sqrt{3}+5) \mathrm{V}$Asked in: AP EAMCET 2021 (23 Aug Shift 2)