As `C=(1/V)Q`, can you say that the capacitor`C` is proportion to the charge `Q`?

Consider the circuit shown where C_(1) = 6 mu F, C_(2) = 3 mu F and V=20 V. Capacitor C_(1) is first charged by closing the switch S_(1). Switch S_(1) is then opened, and the charged capacitor is connected to the uncharged capacitor C_(2) by closing S_(2).

The given graph shows variation of charge .q. versus potential difference .V. for two capacitors C_(1)andC_(2) . Both the capacitors have same plate separation but plate area of C_(2) is greater than that of C_(1) . Which line (A or B) corresponds to C_(1) and why?

Consider the situation shown in (figure) what are the signs of q_1 and q_2 ? If the lines are drawn in proportion to the charge, what is the ratio q_1/q_2 ?

Two capacitors of equal capacitance (C_(1)=C_(2)) are as shown in the figure. Initially, while the swithc is open (as shown) one of the capacitors is uncharged and the other carries charge Q_(0) . The energy stored in the charged capacitor is U_(0). Sometime after the switch is closed, the capacitors C_(1)"and" C_(2) carry charged Q_(1)"and"Q_(2) respectively, the energy stored in the capacitors are U_(1)"and"U_(2) respectively. Which of the following expression is correct?

Shows these variation of charge q for two capacitors C_(1) and C_(2) The two capacitors have the same plate separation. But the plate area of C_(2) is double than that of C_(1) . Which of the lines in the figure corresponds to C_(1) and C_(2) and why? .

Consider a parallel plate capacitor having charge Q. Then,

In the circuit shown, all capacitors are identical. Initially, the switch is open and the capacitor marked C_1 is the only one charged to a value Q_0 . After the switch is closed and the equilibrium is reestablished, the charge on the capacitor marked C_1 is Q. Find the ratio initial change to final charge in capacitor C_1 .

A dielectric slab is inserted between the plates of a capacitor. The charge on the capacitor is Q and the magnitude of the induced charge on each surface of the dielectric is Q' .

The capacitor C_(1) in the figure initially carries a charge q_(0) . When the switch S_(1) and S_(2) are shut, capacitor C_(1) is connected in series to a resistor R and a second capacitor C_(2) , which initially does not carry any charge. If the heat lost in the resistor after a long time of closing the switch is (q_(0)^(2)C_(2))/(kC_(1)(C_(1)+C_(2))) , then Find the value of k .

Two identical capacitors 1 and 2 are connected in series to a batery as shown in figure. Capacitor 2 contains a dielectric slab of dieletric constant k as shown. Q_(1) and Q_(2) are the charges stored in the capacitors. Now the dielectirc slab us removed and the corresponding charges are Q'_(1) and Q'_(2) . Then