Question:
The electric field in a region is given by \(\vec{E} = (Ax + B)\hat{i}\), where \(E\) is in \(\mathrm{NC^{-1}}\) and \(x\) is in metres. The values of constants are \(A = 20\) SI unit and \(B = 10\) SI unit. If the potential at \(x = 1\) is \(V_1\) and that at \(x = -5\) is \(V_2\), then \(V_1 - V_2\) is :-
The electric field in a region is given by \(\vec{E} = (Ax + B)\hat{i}\), where \(E\) is in \(\mathrm{NC^{-1}}\) and \(x\) is in metres. The values of constants are \(A = 20\) SI unit and \(B = 10\) SI unit. If the potential at \(x = 1\) is \(V_1\) and that at \(x = -5\) is \(V_2\), then \(V_1 - V_2\) is :-
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Use \(V_1 - V_2 = -\int_{x_2}^{x_1} E\,dx\) with \(x_2 = -5\) and \(x_1 = 1\).
Updated On: Nov 5, 2026
- \(-520\ \mathrm{V}\)
- \(180\ \mathrm{V}\)
- \(-48\ \mathrm{V}\)
- \(320\ \mathrm{V}\)
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The Correct Option is B
Solution and Explanation
\(E = 20x + 10\).
\(V_1 - V_2 = -\int_{-5}^{1}(20x + 10)\,dx = -\left[10x^2 + 10x\right]_{-5}^{1}\).
At \(x = 1\): \(10 + 10 = 20\). At \(x = -5\): \(250 - 50 = 200\).
\(V_1 - V_2 = -(20 - 200) = 180\ \mathrm{V}\).
Answer: (2) \(180\ \mathrm{V}\)
\(V_1 - V_2 = -\int_{-5}^{1}(20x + 10)\,dx = -\left[10x^2 + 10x\right]_{-5}^{1}\).
At \(x = 1\): \(10 + 10 = 20\). At \(x = -5\): \(250 - 50 = 200\).
\(V_1 - V_2 = -(20 - 200) = 180\ \mathrm{V}\).
Answer: (2) \(180\ \mathrm{V}\)
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