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Electrostatics question

2019 · 9 Apr · Shift 2 · Q54
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Electrostatics question

2019 · 9 Apr · Shift 2 · Q54

JEE MainPhysicsElectrostaticsMCQ+4 / −1
Four point charges –q, +q, +q and –q are placed on y-axis at y = –2d, y = –d, y = +d and y = +2d, respectively. The magnitude of the electric field E at a point on the x-axis at x = D, with D >> d, will behave as :-
  1. A
    E∝1D3E \propto {1 \over D^3}E∝D31​
  2. B
    E∝1DE \propto {1 \over D}E∝D1​
  3. C
    E∝1D4E \propto {1 \over D^4}E∝D41​
  4. D
    E∝1D2E \propto {1 \over D^2}E∝D21​
View written solutionFree

Correct answer: C

  1. Set up the configuration

Charges are placed on the yyy-axis:

  • at y=−2dy=-2dy=−2d: charge −q-q−q
  • at y=−dy=-dy=−d: charge +q+q+q
  • at y=+dy=+dy=+d: charge +q+q+q
  • at y=+2dy=+2dy=+2d: charge −q-q−q

We need the electric field at point P(D,0)P(D,0)P(D,0) on the xxx-axis, where D≫dD \gg dD≫d.

  1. Use symmetry

For a charge at (0,a)(0,a)(0,a), the field at (D,0)(D,0)(D,0) is

E⃗a=14πε0qa(D2+a2)3/2(Di^−aj^)\vec E_a = \frac{1}{4\pi\varepsilon_0}\frac{q_a}{(D^2+a^2)^{3/2}}(D\hat i-a\hat j)Ea​=4πε0​1​(D2+a2)3/2qa​​(Di^−aj^​)

Now pair the charges symmetrically about the origin.

Pair 1: charges at y=±dy=\pm dy=±d

Both are +q+q+q.

  • At y=+dy=+dy=+d:
E⃗+d=14πε0q(D2+d2)3/2(Di^−dj^)\vec E_{+d}=\frac{1}{4\pi\varepsilon_0}\frac{q}{(D^2+d^2)^{3/2}}(D\hat i-d\hat j)E+d​=4πε0​1​(D2+d2)3/2q​(Di^−dj^​)
  • At y=−dy=-dy=−d:
E⃗−d=14πε0q(D2+d2)3/2(Di^+dj^)\vec E_{-d}=\frac{1}{4\pi\varepsilon_0}\frac{q}{(D^2+d^2)^{3/2}}(D\hat i+d\hat j)E−d​=4πε0​1​(D2+d2)3/2q​(Di^+dj^​)

Adding,

E⃗±d=14πε02qD(D2+d2)3/2i^\vec E_{\pm d}=\frac{1}{4\pi\varepsilon_0}\frac{2qD}{(D^2+d^2)^{3/2}}\hat iE±d​=4πε0​1​(D2+d2)3/22qD​i^

The yyy-components cancel.

Pair 2: charges at y=±2dy=\pm 2dy=±2d

Both are −q-q−q.

  • At y=+2dy=+2dy=+2d:
E⃗+2d=14πε0−q(D2+4d2)3/2(Di^−2dj^)\vec E_{+2d}=\frac{1}{4\pi\varepsilon_0}\frac{-q}{(D^2+4d^2)^{3/2}}(D\hat i-2d\hat j)E+2d​=4πε0​1​(D2+4d2)3/2−q​(Di^−2dj^​)
  • At y=−2dy=-2dy=−2d:
E⃗−2d=14πε0−q(D2+4d2)3/2(Di^+2dj^)\vec E_{-2d}=\frac{1}{4\pi\varepsilon_0}\frac{-q}{(D^2+4d^2)^{3/2}}(D\hat i+2d\hat j)E−2d​=4πε0​1​(D2+4d2)3/2−q​(Di^+2dj^​)

Adding,

E⃗±2d=14πε0−2qD(D2+4d2)3/2i^\vec E_{\pm 2d}=\frac{1}{4\pi\varepsilon_0}\frac{-2qD}{(D^2+4d^2)^{3/2}}\hat iE±2d​=4πε0​1​(D2+4d2)3/2−2qD​i^

Again, the yyy-components cancel.

So total field is purely along xxx:

E⃗=14πε02qD[1(D2+d2)3/2−1(D2+4d2)3/2]i^\vec E=\frac{1}{4\pi\varepsilon_0}2qD\left[\frac{1}{(D^2+d^2)^{3/2}}-\frac{1}{(D^2+4d^2)^{3/2}}\right]\hat iE=4πε0​1​2qD[(D2+d2)3/21​−(D2+4d2)3/21​]i^
  1. Approximate for D≫dD\gg dD≫d

Factor out D3D^3D3 from each denominator:

1(D2+a2)3/2=1D3(1+a2D2)−3/2\frac{1}{(D^2+a^2)^{3/2}}=\frac{1}{D^3}\left(1+\frac{a^2}{D^2}\right)^{-3/2}(D2+a2)3/21​=D31​(1+D2a2​)−3/2

Using binomial expansion,

(1+x)−3/2≈1−32x(1+x)^{-3/2}\approx 1-\frac{3}{2}x(1+x)−3/2≈1−23​x

for small xxx.

Thus,

1(D2+d2)3/2≈1D3(1−3d22D2)\frac{1}{(D^2+d^2)^{3/2}}\approx \frac{1}{D^3}\left(1-\frac{3d^2}{2D^2}\right)(D2+d2)3/21​≈D31​(1−2D23d2​)

and

1(D2+4d2)3/2≈1D3(1−32⋅4d2D2)=1D3(1−6d2D2)\frac{1}{(D^2+4d^2)^{3/2}}\approx \frac{1}{D^3}\left(1-\frac{3}{2}\cdot\frac{4d^2}{D^2}\right) =\frac{1}{D^3}\left(1-\frac{6d^2}{D^2}\right)(D2+4d2)3/21​≈D31​(1−23​⋅D24d2​)=D31​(1−D26d2​)

Subtracting,

1(D2+d2)3/2−1(D2+4d2)3/2≈1D3[−3d22D2+6d2D2]=9d22D5\frac{1}{(D^2+d^2)^{3/2}}-\frac{1}{(D^2+4d^2)^{3/2}} \approx \frac{1}{D^3}\left[-\frac{3d^2}{2D^2}+\frac{6d^2}{D^2}\right] =\frac{9d^2}{2D^5}(D2+d2)3/21​−(D2+4d2)3/21​≈D31​[−2D23d2​+D26d2​]=2D59d2​

Therefore,

E≈14πε02qD⋅9d22D5=14πε09qd2D4E \approx \frac{1}{4\pi\varepsilon_0}2qD\cdot \frac{9d^2}{2D^5} =\frac{1}{4\pi\varepsilon_0}\frac{9qd^2}{D^4}E≈4πε0​1​2qD⋅2D59d2​=4πε0​1​D49qd2​

Hence,

E∝1D4E \propto \frac{1}{D^4}E∝D41​
  1. Check options
  • A: E∝1D3E\propto \dfrac{1}{D^3}E∝D31​ ❌
  • B: E∝1DE\propto \dfrac{1}{D}E∝D1​ ❌
  • C: E∝1D4E\propto \dfrac{1}{D^4}E∝D41​ ✅
  • D: E∝1D2E\propto \dfrac{1}{D^2}E∝D21​ ❌
  1. Final answer

The electric field behaves as

E∝1D4E \propto \frac{1}{D^4}E∝D41​

So the correct option is C.

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