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Electrostatics question

2019 · 8 Apr · Shift 2 · Q64
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Electrostatics question

2019 · 8 Apr · Shift 2 · Q64

JEE MainPhysicsElectrostaticsMCQ+4 / −1
The electric field in a region is given by E→=(Ax+B)i∧\mathop E\limits^ \to = \left( {Ax + B} \right)\mathop i\limits^ \wedgeE→​=(Ax+B)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 :-
  1. A
    –520 V
  2. B
    180 V
  3. C
    –48 V
  4. D
    320 V
View written solutionFree

Correct answer: B

  1. Given electric field

    E⃗=(Ax+B)i^\vec E = (Ax+B)\hat iE=(Ax+B)i^

    where: A=20,B=10A=20,\quad B=10A=20,B=10

    So, Ex=20x+10E_x = 20x+10Ex​=20x+10

  2. Relation between electric field and potential

    In one dimension, Ex=−dVdxE_x=-\frac{dV}{dx}Ex​=−dxdV​

    Hence, dV=−Ex dxdV = -E_x\,dxdV=−Ex​dx

    Therefore, V1−V2=V(1)−V(−5)=−∫−51(20x+10) dxV_1 - V_2 = V(1)-V(-5) = -\int_{-5}^{1} (20x+10)\,dxV1​−V2​=V(1)−V(−5)=−∫−51​(20x+10)dx

  3. Evaluate the integral

    ∫(20x+10) dx=10x2+10x\int (20x+10)\,dx = 10x^2 + 10x∫(20x+10)dx=10x2+10x

    So, ∫−51(20x+10) dx=[10x2+10x]−51\int_{-5}^{1} (20x+10)\,dx = \left[10x^2+10x\right]_{-5}^{1}∫−51​(20x+10)dx=[10x2+10x]−51​

    At x=1x=1x=1: 10(1)2+10(1)=10+10=2010(1)^2+10(1)=10+10=2010(1)2+10(1)=10+10=20

    At x=−5x=-5x=−5: 10(25)+10(−5)=250−50=20010(25)+10(-5)=250-50=20010(25)+10(−5)=250−50=200

    Thus, ∫−51(20x+10) dx=20−200=−180\int_{-5}^{1} (20x+10)\,dx = 20-200=-180∫−51​(20x+10)dx=20−200=−180

  4. Now apply the negative sign

    V1−V2=−(−180)=180 VV_1-V_2 = -(-180)=180\text{ V}V1​−V2​=−(−180)=180 V

  5. Option check

    • A: −520 V-520\text{ V}−520 V ❌
    • B: 180 V180\text{ V}180 V ✅
    • C: −48 V-48\text{ V}−48 V ❌
    • D: 320 V320\text{ V}320 V ❌

So the correct answer is Option B.

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