# Doubt in Capacitor

**Ankith_2020**#1

Its given that there exists a filed between the capacitor plates initially, when it was not connected with any battery or anything. This means that the capacitor is having an initial charge, let this charge be q_o.

Now, when the plates are connected with a wire, since the wires will have a resistance, therefore a current will flow through the capacitors by virtue of the leaking of charges through them and this will decay the field between the capacitor plates.

We know that the charge decay equation in case of any capacitor is given by : q(t)=q_oe^{\frac{-t}{RC}}.

Now, we know that the electric field between the plates of any capacitor is given by : E=\frac{q}{A\epsilon_o}

Thus, let the initial field between the capacitor plates be E_o and hence we have : E_o=\frac{q_o}{A\epsilon_o}

Now, the final field at any time interval t is given by : E(t)=\frac{q}{A\epsilon_o}

Given that : E(t)=\frac{1}{3}E_o \implies q(t)=\frac{1}{3}q_o \implies q_oe^{\frac{-t}{RC}}=\frac{1}{3}q_o

\implies R=\frac{t}{Clog_e3}

Given that t=4.4\mu s and C=2.0\mu F, we get : R=\frac{4.4}{2log_e3}

\implies R\approx2\Omega

Thus the answer should be 2\Omega (if integer type is asked, otherwise the exact value is 2.002\Omega).

Do let me know if you have any doubts.

Take a cuboidal section inside the capacitor plates.

The net flux inside this space is zero.Electric field lines are parallel in four of its sides and in two sides there is no electric field(inside capacitor which is conductor and net electric field inside conductor is zero)

So net charge enclosed is zero ,which implies Q2=-Q3.

The electric field inside capacitor in caused only due to facing sides hence answer is B,C.

If you didn’t find my explaination satisfactory do check below images(Source : Cengage physics)

Can u be a bit more explicit in ur statement @Bhuvanitha_2020 ? I didn't get which region u wanna focus on.

**Ankith_2020**#10

Half of Sum of total net charge must appear on outer surface. I am not getting what you want to explain.

**Bhuvanitha_2020**#11

Charges on facing sides forms capacitor.(Equal and opposite sign)

Charges on non facing sides will not produce any electric field in between the space of capacitor plates.

I tried my best to explain ..if you have any doubts pls ask.