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Electrostatics question

2025 · 2 Apr · Shift 1 · Q61
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Electrostatics question

2025 · 2 Apr · Shift 1 · Q61

JEE MainPhysicsElectrostaticsMCQ+4 / −1
A point charge +q+q+q is placed at the origin. A second point charge +9q+9 q+9q is placed at (d,0,0\mathrm{d}, 0,0d,0,0) in Cartesian coordinate system. The point in between them where the electric field vanishes is:
  1. A
    (3d/4,0,0)(3 \mathrm{d} / 4,0,0)(3d/4,0,0)
  2. B
    (d/4,0,0)(\mathrm{d} / 4,0,0)(d/4,0,0)
  3. C
    (4d/3,0,0)(4 \mathrm{d} / 3,0,0)(4d/3,0,0)
  4. D
    (d/3,0,0)(\mathrm{d} / 3,0,0)(d/3,0,0)
View written solutionFree

Correct answer: B

  1. Identify where the net electric field can be zero

For two positive charges, the electric fields due to each charge point away from the charges.

  • To the left of the origin, both fields point in the same direction, so they cannot cancel.
  • To the right of x=dx=dx=d, both fields again point in the same direction, so they cannot cancel.
  • Between the charges, the fields are opposite in direction, so cancellation is possible there.

So let the required point be at xxx on the xxx-axis, where 0<x<d0<x<d0<x<d.

  1. Write magnitudes of the electric fields

Field due to charge +q+q+q at origin:

E1=kqx2E_1 = \frac{kq}{x^2}E1​=x2kq​

Field due to charge +9q+9q+9q at x=dx=dx=d:

E2=k(9q)(d−x)2E_2 = \frac{k(9q)}{(d-x)^2}E2​=(d−x)2k(9q)​

At the null point between them, these magnitudes must be equal:

kqx2=9kq(d−x)2\frac{kq}{x^2} = \frac{9kq}{(d-x)^2}x2kq​=(d−x)29kq​

Cancel kqkqkq:

1x2=9(d−x)2\frac{1}{x^2} = \frac{9}{(d-x)^2}x21​=(d−x)29​
  1. Solve for xxx

Cross-multiplying,

(d−x)2=9x2(d-x)^2 = 9x^2(d−x)2=9x2

Taking positive root since distances are positive,

d−x=3xd-x = 3xd−x=3x d=4xd = 4xd=4x x=d4x = \frac{d}{4}x=4d​

So the electric field vanishes at:

(d4,0,0)\left(\frac{d}{4},0,0\right)(4d​,0,0)
  1. Check the options
  • A: (3d4,0,0)\left(\frac{3d}{4},0,0\right)(43d​,0,0) — incorrect
  • B: (d4,0,0)\left(\frac{d}{4},0,0\right)(4d​,0,0) — correct
  • C: (4d3,0,0)\left(\frac{4d}{3},0,0\right)(34d​,0,0) — outside the interval, incorrect
  • D: (d3,0,0)\left(\frac{d}{3},0,0\right)(3d​,0,0) — incorrect
  1. Comparison with stored answer

Derived answer: B

Stored correct answer: B

They agree.

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