In the adjoining figure- E - 5 V- r - 1- R2 - 4- R1 - R3 - 1- and C - 3- F- The numerical value of the charge on each plate of the capacitor is

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New Capacitance in Dielectric

In the adjoin...

Question

In the adjoining figure, $E = 5 V, r = 1 Ω, R_{2} = 4 Ω, R_{1} = R_{3} = 1 Ω$ and $C = 3 μ F$ . The numerical value of the charge on each plate of the capacitor is

3 μ C

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6 μ C

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12 μ C

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24 μ C

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Solution

The correct option is B $6 μ C$
From figure it is clear that, current will flow only through the branch containing $R_{2}$ .
$∴ l = \frac{E}{R_{2} + r} = \frac{5}{4 + 1} = 1 A$
So, potential difference across $R_{2}; V = l R_{2} = 1 \times 4 = 4 V$
Let $q$ be the charge on each plate of each capacitor, then
$\frac{q}{C} + \frac{q}{C} = 4$
$\Rightarrow \frac{2 q}{C} = 4$
$\Rightarrow q = 2 \times C = 2 \times 3 = 6 μ C$ .

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