Question:
The electric field in a region of space is given by $ \vec{E} = (5\, \text{N/C})\,x\,\hat{i} $. Consider point A on the y-axis at $ y = 5\, \text{m} $ and point B on the x-axis at $ x = 2\, \text{m} $. If the potentials at A and B are $ V_A $ and $ V_B $ respectively, then $ (V_B - V_A) $ is:
The electric field in a region of space is given by $ \vec{E} = (5\, \text{N/C})\,x\,\hat{i} $. Consider point A on the y-axis at $ y = 5\, \text{m} $ and point B on the x-axis at $ x = 2\, \text{m} $. If the potentials at A and B are $ V_A $ and $ V_B $ respectively, then $ (V_B - V_A) $ is:
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Potential difference is negative of line integral of electric field: \( V = -\int \vec{E} \cdot d\vec{r} \)
Updated On: May 20, 2025
- \( -15\, \text{V} \)
- \( 8\, \text{V} \)
- \( -10\, \text{V} \)
- \( -12.5\, \text{V} \)
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The Correct Option is C
Solution and Explanation
Since \( \vec{E} = 5x\, \hat{i} \), the electric field varies with \( x \) only.
So, potential difference:
\[
V_B - V_A = -\int_A^B \vec{E} \cdot d\vec{r}
\]
Point A lies on the y-axis → \( x = 0 \), and point B is on x-axis at \( x = 2 \)
\[
\Rightarrow V_B - V_A = -\int_0^2 5x\, dx = -\left[\frac{5x^2}{2}\right]_0^2
= -\frac{5 \cdot 4}{2} = -10\, \text{V}
\]
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