SATHEE: Electrostatics 5 Question 46

Electrostatics 5 Question 46

48. Two capacitors $A$ and $B$ with capacities $3 μ F$ and $2 μ F$ are charged to a potential difference of $100 V$ and $180 V$ respectively. The plates of the capacitors are connected as shown in the figure with one wire of each capacitor free. The upper plate of $A$ is positive and that of $B$ is negative. An uncharged $2 μ F$ capacitor $C$ with lead wires falls on the free ends to complete the circuit. Calculate

$(1997, 5 M)$

(a) the final charge on the three capacitors and

(b) the amount of electrostatic energy stored in the system before and after completion of the circuit.

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Answer:

Correct Answer: 48. (a) $q_{1} = 90 μ C, q_{2} = 210 μ C, q_{3} = 150 μ C$

(b) (i) $U_{i} = 47.4 mJ$

(ii) $U_{f} = 18 mJ$

Solution:

(a) Charge on capacitor $A$ , before joining with an uncharged capacitor

Similarly, charge on capacitor $B$

$q_{B} = (180) (2) μ C = 360 μ C$

Let $q_{1}, q_{2}$ and $q_{3}$ be the charges on the three capacitors after joining them as shown in figure.

$(q_{1}, q_{2}$ and $q_{3}$ are in microcoulombs)

From conservation of charge

Net charge on plates 2 and 3 before joining $=$ net charge after joining

$∴ 300 = q_{1} + q_{2}$

Similarly, net charge on plates 4 and 5 before joining

$=$ net charge after joining $- 360 = - q_{2} - q_{3}$

or $360 = q_{2} + q_{3}$

Applying Kirchhoff’s second law in closed loop $A B C D A$

$\begin{aligned} \frac{q_{1}}{3} - \frac{q_{2}}{2} + \frac{q_{3}}{2} & = 0 \\ or 2 q_{1} - 3 q_{2} + 3 q_{3} & = 0 \end{aligned}$

Solving Eqs. (i), (ii) and (iii), we get

$\begin{array}{ll} q_{1} & = 90 μ C, q_{2} = 210 μ C \\ and q_{3} & = 150 μ C \end{array}$

(b) (i) Electrostatic energy stored before, completing the circuit

$\begin{matrix} U_{i} = \frac{1}{2} (3 \times 10^{- 6}) (100)^{2} + \frac{1}{2} (2 \times 10^{- 6}) (180)^{2} \\ ∵ U = \frac{1}{2} C V^{2} \\ = 4.74 \times 10^{- 2} J or U_{i} = 47.4 mJ \end{matrix}$

(ii) Electrostatic energy stored after, completing the circuit

$\begin{aligned} U_{f} = & \frac{1}{2} \frac{{(90 \times 10^{- 6})}^{2}}{(3 \times 10^{- 6})} + \frac{1}{2} \frac{{(210 \times 10^{- 6})}^{2}}{(2 \times 10^{- 6})} \\ + \frac{1}{2} \frac{{(150 \times 10^{- 6})}^{2}}{(2 \times 10^{- 6})} U = \frac{1}{2} \frac{q^{2}}{C} \\ = 1.8 \times 10^{- 2} J or U_{f} = 18 mJ \end{aligned}$

Electrostatics 5 Question 46

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Solution:

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Electrostatics 5 Question 46

Answer:

Solution:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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