In the circuit shown (figure1), all the switches are open initially and all capacitors are uncharged. Now first only $S_{1}$ is closed and open when steady state is reached, then $S_{2}$ is closed and open when again steady state is reached, after this $S_{3}$ is closed and open when steady state is reached, this process repeated continuously. After closing sixth switch, the fourth capacitor (8C) is removed and connected (say at t = 0) to an uncharged capacitor of original capacitance C which now has a dielectric (dielectric constant 2) filling half its volume (see figure2), through a resistor R.

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Final charge on capacitor B when steady state reached in figure 2 $\frac{128}{23} C V$

Final charge (when steady state reached in figure 2) on capacitor of capacitance 8C is $\frac{128}{57} C V$

Final charge (when steady state reached in figure 2) on capacitor of capacitance 8C is $\frac{81}{19} C V$

Final charge on capacitor B when steady state reached in figure 2 $\frac{8}{19} C V$

answer is B, C.

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Detailed Solution

After closing $S_{1}$ charge on C = 27 CV
After closing $S_{2}$ charge on $2 C = 2 C . \frac{27 C V}{3 C} = 18 C V$
After closing $S_{3}$ charge on $4 C = 4 C . \frac{18 C V}{6 C} = 12 C V$
After closing $S_{4}$ charge on $8 C = 8 C . \frac{12 C V}{12 C} = 8 C V$
After closing $S_{5}$ charge on $8 C = 8 C . \frac{8 C V}{24 C} = \frac{8 C V}{3}$
Final charge on 8C in figure (2)
$Q = 8 C . \frac{\frac{8 C V}{3}}{8 C + \frac{3 C}{2}} = \frac{\frac{64 C V}{3}}{\frac{19}{2}} = \frac{128 C V}{57}$
Final charge on capacitor B is
$Q = \frac{3 C}{2} . \frac{\frac{8 C V}{3}}{8 C + \frac{3 C}{2}} = \frac{8 C V}{19}$
E at $P = \frac{Δ V}{d} = \frac{16 V}{57 d}$