A charged sphere is connected to a similar uncharged sphere. Then the percentage loss of energy is

Options

A. `50%`
B. `25%`
C. `20%`
D. `10%`

Answer

To solve the problem of finding the percentage loss of energy when a charged sphere is connected to an identical uncharged sphere, we can follow these steps:

### Step-by-Step Solution:

1. **Identify Initial Conditions:**
- Let the charge on the initially charged sphere be \( q \).
- Let the capacitance of each sphere be \( C \).

2. **Calculate Initial Energy:**
- The energy stored in a capacitor is given by the formula:
\[ U = \frac{q^2}{2C} \]
- For the initially charged sphere, the initial energy \( U_i \) is:
\[ U_i = \frac{q^2}{2C} \]

3. **Connect the Spheres:**
- When the charged sphere is connected to the uncharged sphere, the total charge \( q \) will redistribute equally between the two identical spheres.
- After connection, each sphere will have a charge of:
\[ q_f = \frac{q}{2} \]

4. **Calculate Final Energy:**
- The energy stored in each sphere after they are connected is:
\[ U_f = \frac{(q_f)^2}{2C} = \frac{\left(\frac{q}{2}\right)^2}{2C} = \frac{q^2}{4 \cdot 2C} = \frac{q^2}{8C} \]
- Since there are two spheres, the total final energy \( U_f \) is:
\[ U_f = 2 \times \frac{q^2}{8C} = \frac{q^2}{4C} \]

5. **Calculate Energy Loss:**
- The energy loss \( \Delta U \) is given by:
\[ \Delta U = U_i - U_f = \frac{q^2}{2C} - \frac{q^2}{4C} \]
- To simplify this, find a common denominator:
\[ \Delta U = \frac{2q^2}{4C} - \frac{q^2}{4C} = \frac{q^2}{4C} \]

6. **Calculate Percentage Loss of Energy:**
- The percentage loss of energy is calculated using the formula:
\[ \text{Percentage Loss} = \left(\frac{\Delta U}{U_i}\right) \times 100 \]
- Substituting the values we have:
\[ \text{Percentage Loss} = \left(\frac{\frac{q^2}{4C}}{\frac{q^2}{2C}}\right) \times 100 = \left(\frac{1/4}{1/2}\right) \times 100 = \left(\frac{1}{2}\right) \times 100 = 50\% \]

### Final Answer:
The percentage loss of energy is **50%**.