Question 78: Electrostats & Capacitance

Question

In the given circuit, a charge of $$+80$$ $$\mu C$$ is given to the upper plate of the $$4$$ $$\mu F$$ capacitor. Then in the steady state, the charge on the upper plate of the $$3$$ $$\mu F$$ capacitor is

Accepted Answer

Let the charges on the $ 3 \mu \mathrm{F} $ and $ 2 \mu \mathrm{F} $ capacitors be $ q $ and $ q' $ respectively. The charge on the lower plate of the $ 4 \mu \mathrm{F} $ capacitor is $ -80 \mu \mathrm{C} $. The lower plate of the $ 4 \mu \mathrm{F} $ capacitor and the upper plates of the $ 2 \mu \mathrm{F} $ and $ 3 \mu \mathrm{F} $ capacitors form an isolated system. Therefore, the net charge on this system must be zero, which can be represented as :
$ q + q' - 80 \mu \mathrm{C} = 0 \quad 	ext{..........(i)} $
The potentials ($ V = q / C $) across these two capacitors are equal, which gives :
$ \frac{q}{3} = \frac{q'}{2} \quad 	ext{..........(ii)} $
By substituting $ q' $ from equation (ii) into equation (i), we can solve for $ q $ and find :
$ q = 48 \mu \mathrm{C} $

Answer

$$ + \,32\,\mu C$$

Answer

$$ + \,40\,\mu C$$

Answer

$$ + \,48\,\mu C$$

Answer

$$ + \,80\,\mu C$$

Electrostats & Capacitance

In the given circuit, a charge of $+80 \mu C is given to the upper plate of the 4 \mu F capacitor. Then in the steady state, the charge on the upper plate of the 3 \mu F$ capacitor is

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