Question

A parallel plate capacitor has a capacitance of 100pF , a plate area of , and a mica dielectric ($\mathit{k}\mathbf{=}\mathbf{5}\mathbf{.}\mathbf{4}$) completely filling the space between the plates. At 50 V potential difference,(a) Calculate the electric field magnitude E in the mica? (b) Calculate the magnitude of free charge on the plates (c) Calculate the magnitude of induced surface charge on the mica.

Short answer

a) The magnitude of the electric field in the mica is$1.0\times {10}^{4}V/m$ .

b) The magnitude of the free charge on the plates is$5\times {10}^{-9}C$ .

c) The magnitude of the surface charge on the mica is 4.1 nC .

Step-by-step solution

The given data

a) Capacitance of capacitor, C = 100pF

b) Area of the plates,$A=100c{m}^2$

c) Dielectric constant of the mica, k = 5.4

d) Potential difference, V = 50 V

Understanding the concept of the capacitance and charge

The electric field between the plates is the ratio of potential difference between the plates to the separation distance. Using the formula for capacitance, we can calculate the separation distance. By substituting this in the equation of electric field, we can find the electric field. We can calculate the free charge on the plates using the concept of capacitance. From the concept of the electric field in the capacitor, we can find the induced surface charge on mica.

Formulae:

The electric field between the capacitor plates,$E=\frac{V}{d}$ …(i)

The capacitance of a parallel plate capacitor with dielectric,$C=\frac{\kappa {\epsilon }_{0}A}{d}$ …(ii)

The electric field due to net flux on one plate of the capacitor,$E=\frac{q_f}{2{\epsilon }_{0}A}$ …(iii)

The charge stored between the capacitor plates,$q_f=CV$ …(iv)

Where,

${\epsilon }_0=8.85\times {10}^{-12}F.m^{-1}$is permittivity of free space

(a) Calculation of the electric field

Rearranging the equation (ii), we can get the formula to find the separation between the plates as:

$d=\frac{\kappa {\epsilon }_{0}A}{C}$

Substitute his value ofin the equation (i), we can get the magnitude of the electric field by substituting the given data as follows:

$$\begin{gathered} E=\frac{CV}{\kappa {\epsilon }_{0}A} \\ =\frac{\left(50\times {10}^{-12}F\right)\times \left(100V\right)}{5.4\times \left(8.85\times {10}^{-12}F.m^{-1}\right)\times \left(100\times {10}^{-4}{m}^2\right)} \\ =1.0\times {10}^{4}V/m \end{gathered}$$

Hence, the value of the field is$1.0\times {10}^{4}V/m$ .

(b) Calculation of free charge on the plates

We can calculate the free charge on the capacitor, by using the given data in equation (iv) as follows:

$$\begin{gathered} q_f=\left(100\times {10}^{-12}F\right)\times \left(50V\right) \\ =5\times {10}^{-9}C \end{gathered}$$

Hence, the value of the charge is$5\times {10}^{-9}C$ .

(c) Calculation of the surface charge on the mica

The total electric field is due to both the free and induced charges. It can be calculated by finding the sum of electric field due to positive free charge on one plate, electric field due to negative free charge on other plate, the electric field due to induced positive charge on one plate and the electric field due to induced negative charge on other plate.

Thus, the induced surface charge on the mica material is given using equation (iii) as follows:

$$\begin{gathered} E=\frac{q_f}{2{\epsilon }_{0}A}+\frac{q_f}{2{\epsilon }_{0}A}-\frac{q_i}{2{\epsilon }_{0}A}-\frac{q_i}{2{\epsilon }_{0}A} \\ E=\frac{q_f-q_i}{{\epsilon }_{0}A} \\ {q}_i=q_f-{\epsilon }_{0}AE \\ =\left(5\times {10}^{-9}C\right)-\left(8.85\times {10}^{-12}F.m^{-1}\times 100\times {10}^{-4}{m}^2\times 1.0\times {10}^{4}V/m\right) \\ =5\times {10}^{-9}C-8.85\times {10}^{-10}C \\ =4.1\times {10}^{-9}C \\ =4.1nC \end{gathered}$$

Hence, the charge value is 4.1 nC .