In the following circuit, the switch S is closed at t = 0. The charge on the capacitor C1 as a function of time will be given by ( C_eq = C_1C_2 \over C_1 + C_2 )

C_eqE[ 1 - \exp ( - t/RC_eq ) ]

Capacitor PYQ — JEE Main 2018 · 16 Apr · Shift 1 · Q71 with Solution

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Correct answer: C

Interpret the circuit

Since the question gives $C_{eq}=\frac{C_1C_2}{C_1+C_2},$ this means the two capacitors are effectively in series in the charging path through resistor $R$ and source $E$ .

So the capacitor combination behaves like a single equivalent capacitor $C_{eq}$ connected in series with $R$ across the battery $E$ .
Replace the two capacitors by the equivalent capacitor

For capacitors in series, $C_{eq}=\frac{C_1C_2}{C_1+C_2}.$

During charging of an $RC$ circuit, the charge on the equivalent capacitor is $q(t)=C_{eq}E\left(1-e^{-t/(RC_{eq})}\right).$
Relate this to the charge on $C_1$

In a series combination of capacitors, the charge on each capacitor is the same at every instant.

Hence, $q_{C_1}(t)=q(t)=C_{eq}E\left(1-e^{-t/(RC_{eq})}\right).$
Match with the options

This exactly matches: $\boxed{C_{eq}E\left[1-\exp\left(-\frac{t}{RC_{eq}}\right)\right]}$

which is Option C.
Check the other options briefly
- A: uses $C_1E$ and exponent $-tR/C_1$ ; dimensionally incorrect exponent.
- B: uses $C_2E$ instead of common series charge expression.
- D: gives a decaying function starting from $C_{eq}E$ , which is not the charging law from zero initial charge.

Therefore, the correct answer is C.