Question:
C = 100 pF, V = 100 V (given spherical capacitor).
Identical capacitor is touched with this capacitor. If change in total energy is α × 10^-7 J then find α = ?
(Combined capacitance = 200 pF)
C = 100 pF, V = 100 V (given spherical capacitor).
Identical capacitor is touched with this capacitor. If change in total energy is α × 10^-7 J then find α = ?
(Combined capacitance = 200 pF)
Identical capacitor is touched with this capacitor. If change in total energy is α × 10^-7 J then find α = ?
(Combined capacitance = 200 pF)
Show Hint
When capacitors are connected in parallel, their capacitances add. To find the change in energy, calculate the energy before and after the connection using \( E = \frac{1}{2} C V^2 \).
Updated On: Apr 7, 2026
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The Correct Option is A
Solution and Explanation
The energy stored in a capacitor is given by the formula:
\[ E = \frac{1}{2} C V^2 \] Step 1: Initial Energy Calculation.
The initial energy stored in the given spherical capacitor is:
\[ E_{\text{initial}} = \frac{1}{2} \times 100 \times 10^{-12} \times (100)^2
E_{\text{initial}} = \frac{1}{2} \times 100 \times 10^{-12} \times 10^4
E_{\text{initial}} = 0.5 \times 10^{-6} \, \text{J} \]
Step 2: Final Energy Calculation.
When the identical capacitor is connected, the combined capacitance becomes 200 pF. The new total energy is:
\[ E_{\text{final}} = \frac{1}{2} \times 200 \times 10^{-12} \times (100)^2
E_{\text{final}} = \frac{1}{2} \times 200 \times 10^{-12} \times 10^4
E_{\text{final}} = 1 \times 10^{-6} \, \text{J} \]
Step 3: Change in Energy.
The change in energy is:
\[ \Delta E = E_{\text{final}} - E_{\text{initial}} = 1 \times 10^{-6} - 0.5 \times 10^{-6} = 0.5 \times 10^{-6} \, \text{J} \] Now, we are given that the change in total energy is \( \alpha \times 10^{-7} \) J, so:
\[ 0.5 \times 10^{-6} = \alpha \times 10^{-7} \] Solving for \( \alpha \):
\[ \alpha = 5 \] Final Answer: 2.5
\[ E = \frac{1}{2} C V^2 \] Step 1: Initial Energy Calculation.
The initial energy stored in the given spherical capacitor is:
\[ E_{\text{initial}} = \frac{1}{2} \times 100 \times 10^{-12} \times (100)^2
E_{\text{initial}} = \frac{1}{2} \times 100 \times 10^{-12} \times 10^4
E_{\text{initial}} = 0.5 \times 10^{-6} \, \text{J} \]
Step 2: Final Energy Calculation.
When the identical capacitor is connected, the combined capacitance becomes 200 pF. The new total energy is:
\[ E_{\text{final}} = \frac{1}{2} \times 200 \times 10^{-12} \times (100)^2
E_{\text{final}} = \frac{1}{2} \times 200 \times 10^{-12} \times 10^4
E_{\text{final}} = 1 \times 10^{-6} \, \text{J} \]
Step 3: Change in Energy.
The change in energy is:
\[ \Delta E = E_{\text{final}} - E_{\text{initial}} = 1 \times 10^{-6} - 0.5 \times 10^{-6} = 0.5 \times 10^{-6} \, \text{J} \] Now, we are given that the change in total energy is \( \alpha \times 10^{-7} \) J, so:
\[ 0.5 \times 10^{-6} = \alpha \times 10^{-7} \] Solving for \( \alpha \):
\[ \alpha = 5 \] Final Answer: 2.5
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