Electrostatics

JEE Main level, single correct, medium

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Question 201 of 311

The electric field in a region is given by E⃗=(Ax+B)i^,\vec{E}=\left(Ax+B\right)\hat{i}, where EE is in NC−1N{C}^{-1} and xx is in metres. The values of constants are A=20SIA=20SI unit and B=10SIB=10SI unit. If the potential at x=1x=1 is V1{V}_{1} and that at x=−5x=-5 is V2,{V}_{2}, then V1−V2{V}_{1}-{V}_{2} is

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Answer: (C) 180V180V

V1−V2=−∫E.dx{V}_{1}-{V}_{2}=-\int E.dx
=−∫−51(ax+b)dx=-\int _{-5}^{1}\left(ax+b\right)dx
=−[ax22+bx]−51={-\left[\dfrac{a{x}^{2}}{2}+bx\right]}_{-5}^{1}
=−[(a2+b)−(a×252−5b)]=-\left[\left(\dfrac{a}{2}+b\right)-\left(\dfrac{\text{a}\times 25}{2}-5b\right)\right]
=−[202+10−250+50]=-\left[\dfrac{20}{2}+10-250+50\right]
=180V=180V

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