Question

In Fig. 25-59, two parallelplatecapacitors Aand Bare connected in parallel across a 600 Vbattery. Each plate has area$\mathbf{80}\mathbf{.}\mathbf{0}\mathbf{}{\mathbf{cm}}^{\mathbf{2}}$; the plate separations are 3.00 mm. Capacitor Ais filled with air; capacitor Bis filled with a dielectric of dielectric constant k = 2.60. Find the magnitude of the electric field within (a) the dielectric of capacitor Band (b) the air of capacitor A.What are the free charge densities$\sigma$on the higher-potential plate of (c) capacitor Aand (d) capacitor B? (e) What is the induced charge densityon the top surface of the dielectric?

Short answer

1. Magnitude of electric field across capacitor A is $2.0\times {10}^5\frac{\mathrm{V}}{\mathrm{m}}$.
2. Magnitude of electric field across capacitor B is $2.0\times {10}^5\frac{\mathrm{V}}{\mathrm{m}}$.
3. Free charge density on higher potential plate of capacitor A is $1.77\times {10}^{-6}\frac{\mathrm{C}}{{\mathrm{m}}^2}$.
4. Free charge density on higher potential plate of capacitor B is $4.60\times {10}^{-6}\frac{\mathrm{C}}{{\mathrm{m}}^2}$.
5. Induced charge density on the top surface of dielectric is $-2.83\times {10}^{-6}\frac{\mathrm{C}}{{\mathrm{m}}^2}$.

Step-by-step solution

The given data

1. Capacitors A and B are connected in parallel.
2. Potential of a battery, v = 600V
3. Area of the plates,$\mathrm{A}=80.0\times {10}^{-4}{\mathrm{cm}}^2$
4. Separation between the plates,$\mathrm{d}=3\times {10}^{-3}\mathrm{m}$
5. Capacitor A is filled with air,$k_1=1.0$
6. Capacitor B is filled with dielectric,$k_2=2.6$

Understanding the concept of the capacitance, energy density and free charge

The capacitance is the capacity of capacitor to store the charges. The number of charges per unit volume is called as charge density. The stored charges across the plates of the capacitor create potential difference and electric field. The two quantities are dependent on the separation between the two plates.

Capacitor A and capacitor B are connected in parallel with the battery, so, the potential drop across them is the same. Hence, the electric field across them is also the same. But they have different dielectric materials inserted, hence, the free charge density for them is different and the charge induced on the second dielectric material is the only difference in the free charge density for both capacitors’ plates.

Formulae:

The electric field due to the potential difference between the capacitor plates,

$E=V/d$ …(i)

Here, is the electric field, V is the potential difference between the plates, d is the separation between the plates.

The surface charge density from flux theorem due to the dielectric,

$\sigma =k{\epsilon }_{0}E$ …(ii)

Here, is surface charge density, k is the dielectric constant of the medium, ${\epsilon }_0$is the permittivity of the free space, E is the electric field.

(a) Calculation of electric field across A

The electric field across capacitor A can be given using equation (i) as

$$\begin{gathered} {\mathrm{E}}_{\mathrm{A}}=\frac{600\mathrm{V}}{3\times {10}^{-3}\mathrm{m}} \\ =2.0\times {10}^5\frac{\mathrm{V}}{\mathrm{m}} \end{gathered}$$

Hence, the value of the field is $2.0\times {10}^5\frac{\mathrm{V}}{\mathrm{m}}$.

(b) Calculation of electric field across B

Since d and V are same for capacitor B as well, therefore, the magnitude of the electric field across the capacitor across B is $2.0\times {10}^5\frac{\mathrm{V}}{\mathrm{m}}$.

(c) Calculation of free charge density of capacitor A

The free charge density at the higher potential of capacitor A can be calculated using equation (ii) as:

$$\begin{gathered} {\mathrm{\sigma }}_{\mathrm{A}}=1\times (8.85\times {10}^{-12}{\mathrm{C}}^2/\mathrm{N}.{\mathrm{m}}^2)\times (2.0\times {10}^5\mathrm{V}/\mathrm{m}) \\ =1.77\times {10}^{-6}\frac{\mathrm{C}}{{\mathrm{m}}^2} \end{gathered}$$

Hence, the value of the charge density is $1.77\times {10}^{-6}\frac{\mathrm{C}}{{\mathrm{m}}^2}$.

(d) Calculation of free charge density of capacitor B

The free charge density at the higher potential of capacitor B can be calculated using equation (ii) as:

$$\begin{gathered} {\mathrm{\sigma }}_{\mathrm{B}}=2.60\times (8.85\times {10}^{-12}{\mathrm{C}}^2/\mathrm{N}.{\mathrm{m}}^2)\times (2.0\times {10}^5\mathrm{V}/\mathrm{m}) \\ =4.60\times {10}^{-6}\frac{\mathrm{C}}{{\mathrm{m}}^2} \end{gathered}$$

Hence, the value of the charge density is $4.60\times {10}^{-6}\frac{\mathrm{C}}{{\mathrm{m}}^2}$.

(e) Calculation of the induced charge density on top surface of dielectric

The induced charge density on the top surface of dielectric can be given as follows:

$$\begin{gathered} \mathrm{\sigma }\text{'}={\mathrm{\sigma }}_{\mathrm{A}}-{\mathrm{\sigma }}_{\mathrm{B}} \\ =(1.77\times {10}^{-6}\mathrm{c}/{\mathrm{m}}^2-4.60\times {10}^{-6}\mathrm{C}/{\mathrm{m}}^2) \\ =-2.83\times {10}^{-6}\mathrm{C}/{\mathrm{m}}^2 \end{gathered}$$

Hence, the value of the induced charge density is $-2.83\times {10}^{-6}\frac{\mathrm{C}}{{\mathrm{m}}^2}$.