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Electrostatics question

2022 · 29 Jun · Shift 2 · Q59
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Electrostatics question

2022 · 29 Jun · Shift 2 · Q59

JEE MainPhysicsElectrostaticsMCQ+4 / −1
If the electric potential at any point (x, y, z) m in space is given by V = 3x2 volt. The electric field at the point (1, 0, 3) m will be :
  1. A
    3 Vm −-− 1, directed along positive x-axis.
  2. B
    3 Vm −-− 1, directed along negative x-axis.
  3. C
    6 Vm −-− 1, directed along positive x-axis.
  4. D
    6 Vm −-− 1, directed along negative x-axis.
View written solutionFree

Correct answer: D

  1. Given potential

    The electric potential is V=3x2 voltV = 3x^2 \text{ volt}V=3x2 volt

  2. Relation between electric field and potential

    Electric field is the negative gradient of potential: E⃗=−∇V\vec{E} = -\nabla VE=−∇V

    Since VVV depends only on xxx, we have E⃗=−dVdx i^\vec{E} = -\frac{dV}{dx}\,\hat{i}E=−dxdV​i^

  3. Differentiate the potential

    dVdx=ddx(3x2)=6x\frac{dV}{dx} = \frac{d}{dx}(3x^2) = 6xdxdV​=dxd​(3x2)=6x

    Therefore, E⃗=−6x i^\vec{E} = -6x\,\hat{i}E=−6xi^

  4. Evaluate at the point (1,0,3)(1,0,3)(1,0,3)

    At x=1x=1x=1, E⃗=−6(1) i^=−6i^ V m−1\vec{E} = -6(1)\,\hat{i} = -6\hat{i}\ \text{V m}^{-1}E=−6(1)i^=−6i^ V m−1

  5. Interpretation

    Magnitude of electric field: ∣E⃗∣=6 V m−1|\vec{E}| = 6\ \text{V m}^{-1}∣E∣=6 V m−1

    Direction: along the negative x-axis.

  6. Compare with options

    • A: 3 V m−13\,\text{V m}^{-1}3V m−1 along positive xxx-axis ❌
    • B: 3 V m−13\,\text{V m}^{-1}3V m−1 along negative xxx-axis ❌
    • C: 6 V m−16\,\text{V m}^{-1}6V m−1 along positive xxx-axis ❌
    • D: 6 V m−16\,\text{V m}^{-1}6V m−1 along negative xxx-axis ✅

Hence, the correct option is D.

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