Parallel Plate Capacitor: Practice Problem & Solution
Two metallic spheres of radii $1 \text{ cm}$ and $2 \text{ cm}$ are given charges $10^{-2} \text{ C}$ and $5 \times 10^{-2} \text{ C}$ respectively. If they are connected by a conducting wire, the final charge on the smaller sphere is (1995)
Short Solution (Best for NEET):
To solve this problem, we apply the core principles of Parallel Plate Capacitor. Understanding the underlying formula is key to arriving at the correct answer below:
Total charge $Q = 10^{-2} + 5 \times 10^{-2} = 6 \times 10^{-2} \text{ C}$. Capacitances are proportional to radii, so $C_1 : C_2 = 1 : 2$. Charge on the smaller sphere is $q_1 = Q \times \frac{C_1}{C_1 + C_2} = 6 \times 10^{-2} \times \frac{1}{3} = 2 \times 10^{-2} \text{ C}$.
Detailed Solution (Best of Understanding Complex Concepts)
Here is a detailed, step-by-step solution for the question:
Given Data
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Radius of smaller sphere ($r_1$) = $1\text{ cm} = 1 \times 10^{-2}\text{ m}$
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Radius of larger sphere ($r_2$) = $2\text{ cm} = 2 \times 10^{-2}\text{ m}$
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Initial charge on smaller sphere ($q_1$) = $10^{-2}\text{ C}$
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Initial charge on larger sphere ($q_2$) = $5 \times 10^{-2}\text{ C}$
Step 1: Calculate Total Initial Charge ($Q_{\text{total}}$)
Since charge is conserved, the total charge before connecting the wire equals the total charge after connection:
$$Q_{\text{total}} = q_1 + q_2$$
$$Q_{\text{total}} = 10^{-2}\text{ C} + 5 \times 10^{-2}\text{ C} = 6 \times 10^{-2}\text{ C}$$
Step 2: Relation Between Capacitance and Radius
The capacitance $C$ of an isolated conducting sphere of radius $r$ is given by:
$$C = 4\pi\varepsilon_0 r$$
Since $4\pi\varepsilon_0$ is constant, capacitance is directly proportional to radius ($C \propto r$):
$$\frac{C_1}{C_2} = \frac{r_1}{r_2} = \frac{1\text{ cm}}{2\text{ cm}} = \frac{1}{2}$$
Step 3: Common Potential ($V$) After Connection
When the spheres are connected by a conducting wire, charge flows between them until both reach the same common potential $V$.
The common potential $V$ is given by:
$$V = \frac{Q_{\text{total}}}{C_1 + C_2}$$
Step 4: Calculate Final Charge on the Smaller Sphere ($q_1'$)
The final charge on the smaller sphere $q_1'$ is:
$$q_1' = C_1 V = C_1 \left( \frac{Q_{\text{total}}}{C_1 + C_2} \right) = Q_{\text{total}} \left( \frac{C_1}{C_1 + C_2} \right)$$
Substituting $\frac{C_1}{C_2} = \frac{r_1}{r_2}$:
$$q_1' = Q_{\text{total}} \left( \frac{r_1}{r_1 + r_2} \right)$$
Substituting the numerical values:
$$q_1' = (6 \times 10^{-2}\text{ C}) \times \left( \frac{1}{1 + 2} \right)$$
$$q_1' = (6 \times 10^{-2}\text{ C}) \times \frac{1}{3}$$
$$q_1' = 2 \times 10^{-2}\text{ C}$$
Final Answer
The final charge on the smaller sphere is $2 \times 10^{-2}\text{ C}$ (Option 4).
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