A $2 μ F$ 'air filled' and $4 μ F$ 'glass filled' parallel plate capacitors are connected in series with a source of potential difference $V_{0}$ . Plate separation of both capacitors is 0.5 mm. Dielectric strength of air is 3 kV/mm and that of glass is 9 kV/mm. What can be maximum value of $V_{0}$ in kV so that none of the capacitors gets breakdown (Answer up to two decimal places)

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Solution

Maximum potential difference across $2 μ F$ and $4 μ F$ can be calculated by:
$V = E \times d$

$V_{2} = 3 k V / m m \times 0.5 m m = 1.5 k V$

and $V_{4} = 9 k V / m \times 0.5 m m = 4.5 k V$

So maximum charge on $2 μ F$ without breakdown $q_{2} = 2 μ F \times 1.5 k V = 3 m C$ and $q_{4} = 4 μ F \times 4.5 k V = 18 m C$

So $\Rightarrow q = C_{e q} V hence, C_{e q} = \frac{4}{3} μ F$

$3 m C = \frac{4}{3} μ F \times V_{m a x} \Rightarrow V_{m a x} = \frac{9}{4} k V = 2.25 k V$

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