JEE PYQ: Electrostatics Question 42
Question 42 - 2019 (08 Apr Shift 2)
The electric field in a region is given by $\vec{E} = (Ax + B)\hat{i}$, where E is in 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:
(1) 320 V
(2) $-48$ V
(3) 180 V
(4) $-520$ V
Show Answer
Answer: (3)
Solution
$V_1 - V_2 = -\int_{-5}^{1}(20x + 10)dx = -\left[10x^2 + 10x\right]_{-5}^{1} = -[(10+10)-(250-50)] = -[20 - 200] = 180$ V.
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