Answer :

Given,

The area of the capacitor plates, A 96/ϵ_{0}) × 10^{–12} Fm

The distance in between each pairs of plates, d 4mm410^{-3} m

The emf of the connected battery, V 10V

**Formula used**

The schematic representation of distribution of charges when connected to the DC battery is shown in the figure. The plate 2) connected to the positive terminal will be positively charged and the one 4) connected to the negative terminal will be negatively charged.

The oposite charges will be induced in plates 1) and 3), whe the battery is connected as shown.

So, there will be three capacitors that are formed namely, 1-2, 2-3 and 3-4. And they are connected in series arrangement.

Let’s assume that each capacitors has a charge Q, and since they are connected in series, the total charge will also be Q.

Hence the charge, Q

Where,

V Potential difference 10V

C_{eq} Equivalent capacitance of the arrangement.

Now, for series arrangement, we know

And C_{1}, C_{2} and C_{3} are the capacitance of capacitors formed by plates 1-2, 2-3 and 3-4 respectively.

Since area and the separation of all the plates are same,

And we know,

Capacitance of the capacitor,

Where

ε_{o} is the permittivity of the free space

A is the area of the plates of the capacitor

d is the separation between the plates of the capacitor

Substituting the given values in the above equation, we get

Hence, Equivalent capacitance is,

Or,

or,

Or,

Hence, from eqn.1, the charge on each pairs will be,

This is the charge on each side of the plates constituting a capacitor. But the plates connected to the battery has either positive charge or negative charge on both sides, as shown in figure. So, the total charge accumulated in the plates connected to the battery will be two times the above value.

Hence, charge on the plates connected to battery will be 2Q,

Or,

Hence the charge on the specific plates will be ±0.16μC, since one plate is positively charged and the other is negatively charged.

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