In Fig. 25-22, how much charge is stored on the...

In Fig. $25-22$, how much charge is stored on the parallel-plate capacitors by the $10.0 \mathrm{~V}$ battery? One is filled with air, and the other is filled with a dielectric for which $\kappa=3.00$; both capacitors have a plate area of $5.00 \times 10^{-3} \mathrm{~m}^{2}$ and a plate separation of $2.00 \mathrm{~mm}$.

Key Concepts

Stored Charge in a Capacitor

The stored charge on a capacitor is calculated using the formula Q = CV, where C is the capacitance and V is the voltage applied across the plates. This relationship implies that for a given voltage, a capacitor with a higher capacitance will store more charge.

Dielectric Constant

The dielectric constant (?) is a measure of how much a dielectric material can increase the capacitance of a capacitor compared to the capacitance with air or vacuum as the insulator. When a dielectric is inserted between the plates, the capacitance becomes C = ? ??(A/d), thereby increasing the capacitor's ability to store charge.

Battery Voltage

The battery voltage represents the potential difference provided by the battery to the capacitor. It is the driving force that causes charges to accumulate on the capacitor's plates, thus determining how much charge can be stored according to Q = CV.

Parallel Plate Capacitor

A parallel plate capacitor consists of two conductive plates separated by an insulating material (or dielectric). Its capacitance is determined by the area of the plates, the distance between them, and the permittivity of the insulator. The basic formula for a capacitor in a vacuum or air is C = ??(A/d), where ?? is the vacuum permittivity, A is the plate area, and d is the separation between the plates.

Capacitance

Capacitance is the measure of a capacitor's ability to store charge per unit voltage. It is defined by the ratio C = Q/V, meaning the stored charge Q increases with both the capacitance and the voltage V applied across the capacitor.

Transcript

00:01 In this question we've been given a parallel combination of two capacitors.

00:05 So this is how both the capacitors are connected and they're connected with the source of v voltage, which is given as 10 volts.

00:13 And they've also mentioned that this, which is the c1 capacitor, has been filled with a dielectric constant, whereas c2 has air in between the plates.

00:24 So c1 is filled with dielectric constant which has three.

00:30 Value and c2 has normal air inside it.

00:33 We have to find the charge on the capacitors on both the capacitors plates.

00:39 So let's see how we can do this.

00:41 So area of the plates has been given to us as 5 into 10 raised to the power minus 3 meters square whereas the distance between the plates for both the capacitors is same as well as the area.

00:54 So this distance is given to us as 2 into 10 raise to the power minus 3 meters.

01:00 Let's see how we can solve this.

01:02 So, we know that the charge stored, so charge stored in any capacitor is equal to the product of capacitance and voltage.

01:11 So, for both the capacitors, we can write it this way.

01:15 So for the first capacitor, q1 is equal to c1 into v1, whereas for the second capacitor, the charge will be c2 into v2.

01:24 Since both of them are in parallel, the voltage across both of them will be the same.

01:30 Is equal to v2.

01:32 Now let's calculate c1 and c2 and then we'll have our answers.

01:37 So for c1 we know c1 is equal to k epsilon not a by d because this is filled with dielectric and this value comes out to be when we put k as 3, epsilon not as 8 .85 into 10 raised to the power minus 12.

01:54 Area has been given to us as 5 into 10 raised to the power minus 3 and the distance between the plates is 2 into 10 raised to the power minus 3.

02:03 So we can cancel this out and on calculation we will get this answer to be as 66 .375 into 10 raised to the power minus 12.

02:15 So this is the value of capacitor, the first capacitance.

02:21 So i'll just write this clearly...

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