Question 602: Electrostats & Capacitance

Question

Two identical parallel plate air capacitors are connected in series to a battery of emf '$$\mathrm{V}$$'. If one of the capacitor is inserted in liquid of dielectric constant '$$\mathrm{K}$$' then, potential difference of the other capacitor will become

Accepted Answer

Let the capacitance of capacitors are $$\mathrm{C}$$ initially.
So, afterward $$\mathrm{C}_1=\mathrm{kC}$$ and $$\mathrm{C}_2=\mathrm{C}$$
p.d. across capacitor $$\mathrm{C}_2$$ is
$$V_2=\frac{C_2 V}{C_1+C_2}=\frac{k C V}{C+k C}=\frac{k V}{k+1}$$

Answer

$$\frac{\mathrm{K}-1}{\mathrm{KV}}$$

Answer

$$\frac{\mathrm{K}+1}{\mathrm{KV}}$$

Answer

$$\left(\frac{\mathrm{KV}}{\mathrm{K}+1}\right)$$

Answer

$$\frac{\mathrm{KV}}{\mathrm{K}-1}$$

Electrostats & Capacitance

Two identical parallel plate air capacitors are connected in series to a battery of emf '$\mathrm{V}'. If one of the capacitor is inserted in liquid of dielectric constant '\mathrm{K}$' then, potential difference of the other capacitor will become

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