1. Initial charge on the first capacitor

Given:

• $C_1 = 10\,\mu F$
• Initial potential $V_1 = 50\,V$

So the initial charge is $Q_{\text{initial}} = C_1 V_1 = 10\,\mu F \times 50\,V = 500\,\mu C$

2. After connecting to the uncharged capacitor in parallel

Let the second capacitor be $C_2$.

After connection, both capacitors have the same final potential: $V_f = 20\,V$

Since the battery is removed, total charge is conserved.

So, $Q_{\text{initial}} = Q_{\text{final}}$

Final total charge: $Q_{\text{final}} = (C_1 + C_2)V_f$

Thus, $500 = (10 + C_2)\times 20$

3. Solve for $C_2$

$500 = 20(10 + C_2)$ $500 = 200 + 20C_2$ $300 = 20C_2$ $C_2 = 15\,\mu F$

4. Check options

• A: $20\,\mu F$ ✗
• B: $15\,\mu F$ ✓
• C: $10\,\mu F$ ✗
• D: $30\,\mu F$ ✗

Therefore, the correct answer is Option B.