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Electrostatics question

2004 · Shift 0 · Q129
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Electrostatics question

2004 · Shift 0 · Q129

JEE MainPhysicsElectrostaticsMCQ+4 / −1
Four charges equal to -QQQ are placed at the four corners of a square and a charge qqq is at its center. If the system is in equilibrium the value of qqq is
  1. A
    −Q2(1+22)- {Q \over 2}\left( {1 + 2\sqrt 2 } \right)−2Q​(1+22​)
  2. B
    Q4(1+22){Q \over 4}\left( {1 + 2\sqrt 2 } \right)4Q​(1+22​)
  3. C
    −Q4(1+22)- {Q \over 4}\left( {1 + 2\sqrt 2 } \right)−4Q​(1+22​)
  4. D
    Q2(1+22){Q \over 2}\left( {1 + 2\sqrt 2 } \right)2Q​(1+22​)
View written solutionFree

Correct answer: B

  1. Interpret the equilibrium condition

Four identical charges −Q-Q−Q are placed at the corners of a square, and a charge qqq is placed at the center.

For the whole system to be in equilibrium, the net force on each charge must be zero.

  • The charge at the center automatically has zero net force because the four corner charges are symmetrically placed.
  • So we only need to make sure that the net force on any corner charge is zero.
  1. Let the side of the square be aaa

Consider one corner charge, say at corner AAA.

It experiences forces due to:

  1. the two adjacent corner charges,
  2. the diagonally opposite corner charge,
  3. the central charge qqq.

Since all corner charges are −Q-Q−Q, the forces due to the other corner charges are repulsive. To balance them, the central charge must attract the corner charge, so qqq must be positive.

  1. Force on corner charge due to adjacent corner charges

Each adjacent corner is at distance aaa.

So magnitude of force from one adjacent charge is

F1=kQ2a2F_1 = k\frac{Q^2}{a^2}F1​=ka2Q2​

where k=14πε0k = \dfrac{1}{4\pi\varepsilon_0}k=4πε0​1​.

There are two such forces, mutually perpendicular. Their resultant is along the diagonal away from the center, with magnitude

Fadj=F12+F12=2 F1=2 kQ2a2.F_{\text{adj}} = \sqrt{F_1^2 + F_1^2} = \sqrt{2}\,F_1 = \sqrt{2}\,k\frac{Q^2}{a^2}.Fadj​=F12​+F12​​=2​F1​=2​ka2Q2​.
  1. Force due to diagonally opposite corner charge

Distance to the opposite corner is

a2.a\sqrt{2}.a2​.

So the force magnitude is

Fdiag=kQ2(a2)2=kQ22a2.F_{\text{diag}} = k\frac{Q^2}{(a\sqrt{2})^2} = k\frac{Q^2}{2a^2}.Fdiag​=k(a2​)2Q2​=k2a2Q2​.

This force is also along the same diagonal, away from the center.

Thus total outward force on the corner charge due to the other three corner charges is

Fout=2 kQ2a2+kQ22a2.F_{\text{out}} = \sqrt{2}\,k\frac{Q^2}{a^2} + k\frac{Q^2}{2a^2}.Fout​=2​ka2Q2​+k2a2Q2​.
  1. Force due to the central charge

Distance from center to any corner is

a2.\frac{a}{\sqrt{2}}.2​a​.

So magnitude of force between center charge qqq and the corner charge −Q-Q−Q is

Fc=k∣q∣Q(a/2)2=k∣q∣Qa2/2=2k∣q∣Qa2.F_c = k\frac{|q|Q}{(a/\sqrt{2})^2} = k\frac{|q|Q}{a^2/2} = 2k\frac{|q|Q}{a^2}.Fc​=k(a/2​)2∣q∣Q​=ka2/2∣q∣Q​=2ka2∣q∣Q​.

For equilibrium, this must act inward and balance the outward force, so

2kqQa2=2 kQ2a2+kQ22a2.2k\frac{qQ}{a^2} = \sqrt{2}\,k\frac{Q^2}{a^2} + k\frac{Q^2}{2a^2}.2ka2qQ​=2​ka2Q2​+k2a2Q2​.

Since q>0q>0q>0, we can write directly:

2qQ=Q2(2+12).2qQ = Q^2\left(\sqrt{2} + \frac12\right).2qQ=Q2(2​+21​).

Divide by 2Q2Q2Q:

q=Q2(2+12).q = \frac{Q}{2}\left(\sqrt{2} + \frac12\right).q=2Q​(2​+21​).

Now simplify:

q=Q2⋅22+12=Q4(1+22).q = \frac{Q}{2}\cdot \frac{2\sqrt{2}+1}{2} = \frac{Q}{4}(1+2\sqrt{2}).q=2Q​⋅222​+1​=4Q​(1+22​).
  1. Match with the options
q=Q4(1+22)q = \frac{Q}{4}(1+2\sqrt{2})q=4Q​(1+22​)

which corresponds to Option B.

  1. Comparison with stored correct answer

Stored correct answer: B

Our derived answer: B

They match.

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