Question

v-q graph for capacitor​

Answer 1

Answer:

1 is the most accurate answer but graph will never tip down at the end , slope tends to become zero at infinity.

Explanation:

Let us assume a case of some resistance in the wires during connection of the capacitor plates to the battery.

Emf of the battery is E

Capacitance of the capacitor is C

resistance of the wires is R

Now , using kirchoff loop rule

E - iR - q/c =0

since I=dq/dt

The above equation is a first order variable separable .

After solving the equation , and placing the boundary conditions ( at t=0 , charge is zero)

the solution is

q=CE( 1 - e^(-t/RC))

we also know that at any instant , potential difference across a capacitor is q/c

V=q/c= E(1 - e^(-t/RC))

since q/v is constant

q=CV the plot is a straight line