JEE Challenger
More from Electrostatic Potential and Capacitance
PhyMedMain 2026 • 6th Apr - Shift 2

Energy Loss on Touching Identical Charged and Uncharged Spheres

A sphere of capacitance 100 pF100\text{ pF} is charged to a potential of 100 V100\text{ V}. Another identical uncharged metal sphere is brought in contact with the charged sphere, then the change in the total energy stored on these spheres, when they touch is α×10−7 J\alpha \times 10^{-7}\text{ J}. The value of α\alpha is ________. (combined capacitance of spheres is 200 pF200\text{ pF})

Options

A

5

B

52\frac{5}{2}

Correct
C

72\frac{7}{2}

D

92\frac{9}{2}

Topics & Concepts

Electrostatic Potential and CapacitanceCapacitorsElectric Potential

Step-by-Step Solution

To find the value of α\alpha, we analyze the initial and final electrostatic potential energy of the system of two spheres.

1. Initial State

  • Capacitance of the first sphere, C1=100 pF=100×10−12 F=10−10 FC_1 = 100\text{ pF} = 100 \times 10^{-12}\text{ F} = 10^{-10}\text{ F}
  • Potential of the first sphere, V1=100 VV_1 = 100\text{ V}
  • The second identical sphere is uncharged: C2=100 pFC_2 = 100\text{ pF} and V2=0 VV_2 = 0\text{ V}

The total initial energy UiU_i stored in the system is given by: Ui=12C1V12+12C2V22U_i = \frac{1}{2} C_1 V_1^2 + \frac{1}{2} C_2 V_2^2

Substituting the given values: Ui=12×(100×10−12 F)×(100 V)2+0U_i = \frac{1}{2} \times (100 \times 10^{-12}\text{ F}) \times (100\text{ V})^2 + 0 Ui=12×10−10×104=5×10−7 JU_i = \frac{1}{2} \times 10^{-10} \times 10^4 = 5 \times 10^{-7}\text{ J}

2. Final State (After Contact)

When the two identical spheres are brought into contact, charge flows between them until both reach a common potential VV.

The common potential VV is given by: V=C1V1+C2V2C1+C2=100×100+0100+100=50 VV = \frac{C_1 V_1 + C_2 V_2}{C_1 + C_2} = \frac{100 \times 100 + 0}{100 + 100} = 50\text{ V}

The combined capacitance of the system is: Ctotal=C1+C2=100 pF+100 pF=200 pF=200×10−12 FC_{\text{total}} = C_1 + C_2 = 100\text{ pF} + 100\text{ pF} = 200\text{ pF} = 200 \times 10^{-12}\text{ F}

The final total energy UfU_f stored in the system is: Uf=12CtotalV2U_f = \frac{1}{2} C_{\text{total}} V^2 Uf=12×(200×10−12 F)×(50 V)2U_f = \frac{1}{2} \times (200 \times 10^{-12}\text{ F}) \times (50\text{ V})^2 Uf=100×10−12×2500=2.5×10−7 JU_f = 100 \times 10^{-12} \times 2500 = 2.5 \times 10^{-7}\text{ J}

3. Change in Total Energy (Energy Loss)

The formula for the energy lost during the redistribution of charge can also be directly written as: ΔU=12C1C2C1+C2(V1−V2)2\Delta U = \frac{1}{2} \frac{C_1 C_2}{C_1 + C_2} (V_1 - V_2)^2

Calculating the loss in energy ΔU\Delta U: ΔU=Ui−Uf=5×10−7 J−2.5×10−7 J=2.5×10−7 J=52×10−7 J\Delta U = U_i - U_f = 5 \times 10^{-7}\text{ J} - 2.5 \times 10^{-7}\text{ J} = 2.5 \times 10^{-7}\text{ J} = \frac{5}{2} \times 10^{-7}\text{ J}

4. Conclusion

Comparing the calculated change in energy with the given expression ΔU=α×10−7 J\Delta U = \alpha \times 10^{-7}\text{ J}: α=52\alpha = \frac{5}{2}

Thus, the correct option is B.

Energy Loss on Touching Identical Charged and Uncharged Spheres | Physics PYQ Solution - JEE Challenger