AP EAMCET · PHYSICS · Electrostatics
A uniform electric field of \(500 \mathrm{Vm}^{-1}\) is directed at \(30^{\circ}\) with the positive \(X\)-axis as shown in figure. The potential difference \(\left(V_B-V_A\right)\) if \(O A=3 \mathrm{~m}\) and \(O B=5 \mathrm{~m}\) is
- A \(-250(3 \sqrt{3}+5) \vee\)
- B \(250(3 \sqrt{3}+5) \vee\)
- C \(-250(3+5 \sqrt{3}) V\)
- D \(250(3+5 \sqrt{3}) V\)
Answer & Solution
Correct Answer
(A) \(-250(3 \sqrt{3}+5) \vee\)
Step-by-step Solution
Detailed explanation
Given, Electric field \(E_0=500 \mathrm{Vm}^{-1}\) \(\begin{aligned} \mathbf{E} & =\left(E_0 \cos \theta \hat{\mathbf{i}}+E_0 \sin \theta \hat{\mathbf{j}}\right) \mathrm{Vm}^{-1} \\ O A & =3 \mathrm{~cm} \text { and } O B=5 \mathrm{~m}\end{aligned}\) Now, displacement vector,…
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