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Question: In a certain region, electric field E exists along the x-axis which is uniform. Given $AB = 2\sqrt{3...

In a certain region, electric field E exists along the x-axis which is uniform. Given AB=23AB = 2\sqrt{3} m and BC=4BC = 4 m. Points, A, B, and C are in XY plane. Find the potential difference VA−VBV_A - V_B between the points A and B.

A

3E

B

4E

C

E

D

2E

Answer

3E

Explanation

Solution

The potential difference between two points A and B in a uniform electric field E⃗\vec{E} is given by VA−VB=E⃗⋅(r⃗B−r⃗A)=E⃗⋅r⃗ABV_A - V_B = \vec{E} \cdot (\vec{r}_B - \vec{r}_A) = \vec{E} \cdot \vec{r}_{AB}, where r⃗AB\vec{r}_{AB} is the displacement vector from A to B. The electric field is given as E⃗=Ei^\vec{E} = E \hat{i}. The length of the line segment AB is 232\sqrt{3} m, and it makes an angle of 30∘30^\circ below the horizontal (positive x-axis). The displacement vector from A to B is r⃗AB=(23cos⁡(−30∘))i^+(23sin⁡(−30∘))j^=(23⋅32)i^+(23⋅(−12))j^=3i^−3j^\vec{r}_{AB} = (2\sqrt{3} \cos(-30^\circ)) \hat{i} + (2\sqrt{3} \sin(-30^\circ)) \hat{j} = (2\sqrt{3} \cdot \frac{\sqrt{3}}{2}) \hat{i} + (2\sqrt{3} \cdot (-\frac{1}{2})) \hat{j} = 3 \hat{i} - \sqrt{3} \hat{j}. The potential difference is VA−VB=E⃗⋅r⃗AB=(Ei^)⋅(3i^−3j^)=E(3)+E(−3)(0)=3EV_A - V_B = \vec{E} \cdot \vec{r}_{AB} = (E \hat{i}) \cdot (3 \hat{i} - \sqrt{3} \hat{j}) = E(3) + E(-\sqrt{3})(0) = 3E.