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Electrostatics question

2025 · 24 Jan · Shift 2 · Q58
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  5. /2025 · 24 Jan · Shift 2 · Q58

Electrostatics question

2025 · 24 Jan · Shift 2 · Q58

JEE MainPhysicsElectrostaticsMCQ+4 / −1
A small uncharged conducting sphere is placed in contact with an identical sphere but having 4×10−8C4 \times 10^{-8} \mathrm{C}4×10−8C charge and then removed to a distance such that the force of repulsion between them is 9×10−3 N9 \times 10^{-3} \mathrm{~N}9×10−3 N. The distance between them is (Take 14πϵo\frac{1}{4 \pi \epsilon_{\mathrm{o}}}4πϵo​1​ as 9×1099 \times 10^99×109 in SI units)
  1. A
    1 cm
  2. B
    2 cm
  3. C
    4 cm
  4. D
    3 cm
View written solutionFree

Correct answer: B

  1. Charge sharing on identical conducting spheres

When two identical conducting spheres are brought into contact, the total charge gets shared equally.

Initial charges:

  • Sphere 1: 000
  • Sphere 2: 4×10−8 C4 \times 10^{-8}\,\text{C}4×10−8C

Total charge: Qtotal=4×10−8 CQ_{\text{total}} = 4 \times 10^{-8}\,\text{C}Qtotal​=4×10−8C

Since the spheres are identical, after contact each sphere gets: q=4×10−82=2×10−8 Cq = \frac{4 \times 10^{-8}}{2} = 2 \times 10^{-8}\,\text{C}q=24×10−8​=2×10−8C

  1. Apply Coulomb's law

After separation, the repulsive force between them is given as: F=9×10−3 NF = 9 \times 10^{-3}\,\text{N}F=9×10−3N

Coulomb's law: F=14πϵ0q2r2F = \frac{1}{4\pi\epsilon_0}\frac{q^2}{r^2}F=4πϵ0​1​r2q2​

Substitute the values: 9×10−3=9×109⋅(2×10−8)2r29 \times 10^{-3} = 9 \times 10^9 \cdot \frac{(2 \times 10^{-8})^2}{r^2}9×10−3=9×109⋅r2(2×10−8)2​

  1. Simplify

First, (2×10−8)2=4×10−16(2 \times 10^{-8})^2 = 4 \times 10^{-16}(2×10−8)2=4×10−16

So, 9×10−3=9×109⋅4×10−16r29 \times 10^{-3} = 9 \times 10^9 \cdot \frac{4 \times 10^{-16}}{r^2}9×10−3=9×109⋅r24×10−16​

9×10−3=36×10−7r29 \times 10^{-3} = \frac{36 \times 10^{-7}}{r^2}9×10−3=r236×10−7​

9×10−3=3.6×10−6r29 \times 10^{-3} = \frac{3.6 \times 10^{-6}}{r^2}9×10−3=r23.6×10−6​

Now solve for r2r^2r2: r2=3.6×10−69×10−3r^2 = \frac{3.6 \times 10^{-6}}{9 \times 10^{-3}}r2=9×10−33.6×10−6​

r2=0.4×10−3=4×10−4r^2 = 0.4 \times 10^{-3} = 4 \times 10^{-4}r2=0.4×10−3=4×10−4

Therefore, r=4×10−4=2×10−2 mr = \sqrt{4 \times 10^{-4}} = 2 \times 10^{-2}\,\text{m}r=4×10−4​=2×10−2m

r=0.02 m=2 cmr = 0.02\,\text{m} = 2\,\text{cm}r=0.02m=2cm

  1. Match with options

Thus the correct option is: B: 2 cm\boxed{\text{B: } 2\,\text{cm}}B: 2cm​

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