In the given circuit, charge $\mathrm{Q}_{2}$ on the $2 \mu \mathrm{~F}$ capacitor changes as C is varied from $1 \mu \mathrm{~F}$ to $3 \mu \mathrm{~F}$. $\mathrm{Q}_{2}$ as a function of 'C' is given properly by: (figures are drawn schematically and are not to scale)

JEE Main · 2015 · 04 Apr 2015 · Offline · Capacitance

In the given circuit, charge

Q_{2}

on the

2 μ F

capacitor changes as C is varied from

1 μ F

3 μ F

Q_{2}

as a function of 'C' is given properly by: (figures are drawn schematically and are not to scale)

Answer: (A)

Let the potential of the node joining the three capacitors be

V

. Charge conservation there gives

C (E - V) = 1 \cdot V + 2 \cdot V

, so

V = \frac{C E}{3 + C}

and the charge on the

2 μ F

capacitor is

Q_{2} = 2 V = \frac{2 C E}{3 + C}

. As

C

grows from

1 μ F

3 μ F

Q_{2}

rises from

0.5 E

E

; since

\frac{d Q _{2}}{d C} = \frac{6 E}{( 3 + C ) ^{2}}

shrinks as

C

increases, the curve goes up but is concave down, bending over and flattening (it would approach

2 E

only as

C \to \infty

). The correct graph is the one that rises and is concave down between these two values of

C

JEE Main · 2015 · 04 Apr 2015 · Offline · Single correct option