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Electrostatics question

2023 · 13 Apr · Shift 2 · Q56
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Electrostatics question

2023 · 13 Apr · Shift 2 · Q56

JEE MainPhysicsElectrostaticsMCQ+4 / −1
A 10 μC10 ~\mu \mathrm{C}10 μC charge is divided into two parts and placed at 1 cm1 \mathrm{~cm}1 cm distance so that the repulsive force between them is maximum. The charges of the two parts are:
  1. A
    9 μC,1 μC9 ~\mu\mathrm{C}, 1 ~\mu \mathrm{C}9 μC,1 μC
  2. B
    5 μC,5 μC5 ~\mu\mathrm{C}, 5 ~\mu \mathrm{C}5 μC,5 μC
  3. C
    8 μC,2 μC8 ~\mu\mathrm{C}, 2 ~\mu \mathrm{C}8 μC,2 μC
  4. D
    7 μC,3 μC7 ~\mu\mathrm{C}, 3 ~\mu \mathrm{C}7 μC,3 μC
View written solutionFree

Correct answer: B

  1. Let the two parts of the charge be qqq and (10−q)(10-q)(10−q) in units of μC\mu\text{C}μC.

  2. Use Coulomb's law

Since the distance between them is fixed at r=1 cmr=1\text{ cm}r=1 cm, the repulsive force is

F=kq(10−q)r2F = k\frac{q(10-q)}{r^2}F=kr2q(10−q)​

where kkk and rrr are constants.

So, to maximize FFF, we need to maximize the product

q(10−q)q(10-q)q(10−q)
  1. Form the function
f(q)=q(10−q)=10q−q2f(q)=q(10-q)=10q-q^2f(q)=q(10−q)=10q−q2

This is a downward opening parabola, so its maximum occurs at the vertex.

  1. Find the maximum

Differentiate:

dfdq=10−2q\frac{df}{dq}=10-2qdqdf​=10−2q

Set it equal to zero:

10−2q=010-2q=010−2q=0 q=5q=5q=5

Then the other charge is

10−q=510-q=510−q=5
  1. Therefore, the force is maximum when the charge is divided equally.

So the two charges are:

5 μC and 5 μC5~\mu\text{C} \text{ and } 5~\mu\text{C}5 μC and 5 μC
  1. Check options
  • A: 9,19,19,1 gives product 999
  • B: 5,55,55,5 gives product 252525
  • C: 8,28,28,2 gives product 161616
  • D: 7,37,37,3 gives product 212121

Maximum product is for Option B.

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