SATHEE: Electrostatics 5 Question 4

Electrostatics 5 Question 4

4. Figure shows charge $(q)$ versus voltage $(V)$ graph for series and parallel combination of two given capacitors. The capacitances are

(Main 2019, 10 April I)

(a) $60 μ F$ and $40 μ F$

(b) $50 μ F$ and $30 μ F$

(d) $40 μ F$ and $10 μ F$

5 The parallel combination of two air filled parallel plate capacitors of capacitance $C$ and $n C$ is connected to a battery of voltage, $V$ . When the capacitors are fully charged, the battery is removed and after that a dielectric material of dielectric constant $K$ is placed between the two plates of the first capacitor. The new potential difference of the combined system is

(a) $\frac{(n + 1) V}{(K + n)}$

(b) $\frac{n V}{K + n}$

(d) $\frac{V}{K + n}$

(Main 2019, 9 April II)

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Answer:

Correct Answer: 4. (b)

Solution:

In the given figure,

Slope of $O A >$ Slope of $O B$

Since, we know that, net capacitance of parallel combination $>$ net capacitance of series combination

$∴$ Parallel combination’s capacitance,

$C_{P} = C_{1} + C_{2} = \frac{500 μ C}{10 V} = 50 μ F$

Series combination’s capacitance,

$C_{S} = \frac{C_{1} C_{2}}{C_{1} + C_{2}} = \frac{80 μ C}{10 V} = 8 μ F$

$\begin{aligned} C_{1} C_{2} & = 8 \times (C_{1} + C_{2}) = 8 \times 50 μ F \\ = 400 μ F [using Eq. (i)] \end{aligned}$

From Eqs. (i) and (iii), we get

$\begin{aligned} C_{1} = 50 - C_{2} \\ and C_{1} C_{2} = 400 \\ \Rightarrow C_{2} (50 - C_{2}) = 400 \\ \Rightarrow 50 C_{2} - C_{2}^{2} = 400 \\ or C_{2}^{2} - 50 C_{2} + 400 = 0 \\ \Rightarrow C_{2} = \frac{+ 50 \pm \sqrt{2500 - 1600}}{2} = \frac{+ 50 \pm 30}{2} \end{aligned}$