Question 604: Electrostats & Capacitance

Question

A condenser of capacity '$$\mathrm{C}_1$$' is charged to potential '$$\mathrm{V}_1$$' and then disconnected. Uncharged capacitor of capacity '$$\mathrm{C}_2$$' is connected in parallel with '$$\mathrm{C}_1$$'. The resultant potential '$$\mathrm{V}_2$$' is

Accepted Answer

The charge stored by the charged condenser is
$$\mathrm{Q}=\mathrm{C}_1 \mathrm{~V}_1$$
When the uncharged condenser is connected in parallel, the effective capacitance becomes $$\left(C_1+C_2\right)$$
Hence the potential $$V_2=\frac{Q}{C_1+C_2}=\frac{C_1 V_1}{C_1+C_2}$$

Answer

$$\frac{V_1 C_2}{C_1}$$

Answer

$$\frac{\mathrm{C}_2}{\mathrm{C}_1+\mathrm{C}_2}$$

Answer

$$\frac{C_1 V_1}{C_2}$$

Answer

$$\frac{C_1 V_1}{C_1+C_2}$$

Electrostats & Capacitance

A condenser of capacity '$\mathrm{C}_1' is charged to potential '\mathrm{V}_1' and then disconnected. Uncharged capacitor of capacity '\mathrm{C}_2' is connected in parallel with '\mathrm{C}_1'. The resultant potential '\mathrm{V}_2$' is

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