Question

A sphere of capacitance \(100\,\text{pF}\) is charged to a potential of \(100\,\text{V}\). 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 \(\alpha \times 10^{-7}\,\text{J}\). The value of \(\alpha\) is _____. (combined capacitance of spheres is \(200\,\text{pF}\)).

• A. \(5\)
• B. \( \frac{5}{2} \)
• C. \( \frac{7}{2} \)
• D. \( \frac{9}{2} \)

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
We need to calculate the energy dissipated (change in energy) when charge is shared between a charged capacitor and an identical uncharged one.
Step 2: Key Formula or Approach:
Energy stored in a capacitor \(U = \frac{1}{2} C V^2\).
The loss in energy during redistribution of charge is:
\[ \Delta U = \frac{1}{2} \frac{C_1 C_2 (V_1 - V_2)^2}{(C_1 + C_2)} \]
Step 3: Detailed Explanation:
Initial conditions:
\(C_1 = 100 \text{ pF}\), \(V_1 = 100 \text{ V}\).
\(C_2 = 100 \text{ pF}\), \(V_2 = 0 \text{ V}\).
Calculating energy loss:
\[ \Delta U = \frac{1}{2} \frac{(100 \times 10^{-12}) (100 \times 10^{-12}) (100 - 0)^2}{(100 + 100) \times 10^{-12}} \]
\[ \Delta U = \frac{1}{2} \frac{100 \times 100 \times 10^{-24} \times 10^4}{200 \times 10^{-12}} \]
\[ \Delta U = \frac{1}{2} \frac{10^8 \times 10^{-24}}{2 \times 10^{-10}} = \frac{10^{-16}}{4 \times 10^{-10}} \]
\[ \Delta U = 0.25 \times 10^{-6} \text{ J} = 2.5 \times 10^{-7} \text{ J} \]
Comparing with the given form \(\alpha \times 10^{-7}\):
\[ \alpha = 2.5 = \frac{5}{2} \]
Step 4: Final Answer:
The value of \(\alpha\) is \(\frac{5}{2}\).