Question 566: Electrostats & Capacitance

Question

A parallel combination of two capacitors of capacities '$$2 ~\mathrm{C}$$' and '$$\mathrm{C}$$' is connected across $$5 \mathrm{~V}$$ battery. When they are fully charged, the charges and energies stored in them be '$$\mathrm{Q}_1$$', '$$Q_2$$' and '$$E_1$$', '$$E_2$$' respectively. Then $$\frac{E_1-E_2}{Q_1-Q_2}$$ in $$\mathrm{J} / \mathrm{C}$$ is (capacity is in Farad, charge in Coulomb and energy in J)

Accepted Answer

We know, $$\mathrm{Q}=\mathrm{C} . \mathrm{V}$$
$$\begin{array}{ll}
	herefore \quad Q_1=10 \mathrm{C} 	ext { and } \mathrm{Q}_2=5 \mathrm{C} & \
	ext { Energy stored, } \mathrm{E}=\frac{1}{2} \mathrm{CV}^2 \
	herefore  \quad \mathrm{E}_1=\frac{1}{2} \mathrm{C}_1 \mathrm{~V}^2=\frac{1}{2} 	imes 2 \mathrm{C} 	imes 25 =25 \mathrm{~J} \
 	ext { Similarly, } \mathrm{E}_2=\frac{1}{2} \mathrm{C}_2 \mathrm{~V}^2=\frac{1}{2} 	imes \mathrm{C} 	imes 25 =12.5~\mathrm{J}\
	herefore \quad \frac{\mathrm{E}_1-\mathrm{E}_2}{\mathrm{Q}_1-\mathrm{Q}_2}=\frac{12.5}{5}=\frac{5}{2}
\end{array}$$

Answer

$$\frac{5}{4}$$

Answer

$$\frac{4}{5}$$

Answer

$$\frac{5}{2}$$

Answer

$$\frac{2}{5}$$

Electrostats & Capacitance

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