A very small sphere of mass 60 g having a charge q is held at a height of 7m vertically above the center of a fixed conducting sphere of radius 1 m, carrying an equal charge q. When released, it falls until it is repelled back just before it comes in contact with the sphere as shown in given figure. Find the charge? (g = 10 m/s2)

F3 Savita Engineering 28-6-22 D6

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  1. \(25 \sqrt{1.5}\)
  2. \(20 \sqrt{1.5}\)
  3. \(20 \sqrt{3}\)
  4. 21.6 μC

Answer (Detailed Solution Below)

Option 4 : 21.6 μC
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Detailed Solution

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Concept:

  • Here, we will use the concept of conservation of energy.
Total energy associated with system initially  =  Total energy associated with system finally
  • Total energy means the total kinetic energy of the system plus the total potential energy of the system.

Calculation:

Given

Mass of charge q = 60g = 0.06 kg 

The distance between the charge is r = 7m

The electric potential energy stored initially,

\(P.E. = \frac{1}{4\pi \epsilon_0}\frac{q.q}{7}\)

The gravitational potential energy by taking the center of the sphere as the reference,

\(P.E. = 7mg\)

The electric potential energy is stored finally,

\(P.E. = \frac{1}{4\pi \epsilon_0}\frac{q.q}{1}\)

The gravitational potential energy by taking the center of the sphere as the reference finally,,

\(P.E. = 1mg\)

Note that here the kinetic energy initially and finally will be zero.

\(\frac{1}{4\pi \epsilon_0}\frac{q.q}{7} + mg×7 = \frac{1}{4\pi \epsilon_0 }\frac{q^2}{1} + mg× 1\)

\(\frac{q^2}{4\pi \epsilon_0}= 7mg\)

\(\frac{q^2}{4\pi \epsilon_0}= 7\times 0.060\times 10\)

k = 9 × 10 9 N ⋅ m 2 /C 2

\(⇒ q^2= \frac{ 7\times 0.060\times 10}{9\times 10^9 } \)

\(⇒ q = 2.16 \times 10^{-5}\ C\)

⇒ q = 21.6 μC

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