Let's analyze the problem step-by-step.
1. Initial Setup:
- We have a capacitor of capacitance [tex]\( C \)[/tex] charged to a voltage [tex]\( V \)[/tex].
- This charged capacitor has an initial charge [tex]\( Q \)[/tex] given by: [tex]\[ Q = C \times V \][/tex]
2. Connecting in Parallel:
- We connect this charged capacitor in parallel with an identical uncharged capacitor (also of capacitance [tex]\( C \)[/tex]).
- When capacitors are connected in parallel, the total capacitance [tex]\( C_{\text{total}} \)[/tex] is the sum of the individual capacitances. Therefore: [tex]\[ C_{\text{total}} = C + C = 2C \][/tex]
3. Charge Redistribution:
- The initial charge [tex]\( Q \)[/tex] on the charged capacitor will redistribute between the two capacitors when they are connected in parallel.
- The total charge in the system remains the same (conservation of charge), which is: [tex]\[ Q = CV \][/tex]
4. Voltage Across Capacitors:
- In a parallel connection, the voltage across each capacitor will be the same. Let's call this common voltage [tex]\( V_{\text{common}} \)[/tex].
- The total charge is now shared between the two capacitors, which both have the same voltage [tex]\( V_{\text{common}} \)[/tex]. The total charge [tex]\( Q \)[/tex] is now: [tex]\[ Q = (2C) \times V_{\text{common}} \][/tex]
5. Solving for [tex]\( V_{\text{common}} \)[/tex]:
- We can use the initial total charge equation: [tex]\[ CV = 2C \times V_{\text{common}} \][/tex]
- Solving for [tex]\( V_{\text{common}} \)[/tex]: [tex]\[ V_{\text{common}} = \frac{CV}{2C} = \frac{V}{2} \][/tex]
6. Charge on Each Capacitor:
- With [tex]\( V_{\text{common}} = \frac{V}{2} \)[/tex], we can find the charge on each capacitor using [tex]\( Q = C \times V_{\text{common}} \)[/tex].
- Therefore, the charge on each capacitor is: [tex]\[ Q_{\text{each}} = C \times \frac{V}{2} = \frac{1}{2} CV \][/tex]
Answer:
The charge on each capacitor, after they are connected in parallel, is [tex]\( \frac{1}{2} CV \)[/tex].
The correct answer is (b) [tex]\(\frac{1}{2} CV\)[/tex].
