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JEE Main 2019 (8 Apr, Shift 2) · Electric Potential & Equipotentials

Electric Potential & Equipotentials question from JEE Main 2019 (8 Apr, Shift 2)

Q97JEE Main 2019 · 8 APR · SHIFT 2Medium

Electric Potential & Equipotentials

The electric field in a region is given by E⃗=(Ax+B)i^\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 V1 and that at x = –5 is V2, then V1 – V2 is :-

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Answer: (b)

  1. V1−V2=−∫−51(20x+10) dx=−[10x2+10x]−51V_1 - V_2 = -\int_{-5}^{1}(20x + 10)\,dx = -\left[10x^2 + 10x\right]_{-5}^{1}
  2. =−[(10+10)−(250−50)]=−(20−200)=180 V= -[(10 + 10) - (250 - 50)] = -(20 - 200) = 180\,\text{V}
  3. So the answer is (b) 180 V.
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