Question

A parallel plate capacitor has a capacitance of \(120 .\) pF and a plate area of \(100 . \mathrm{cm}^{2}\). The space between the plates is filled with mica whose dielectric constant is \(5.40 .\) The plates of the capacitor are kept at \(50.0 \mathrm{~V}\) a) What is the strength of the electric field in the mica? b) What is the amount of free charge on the plates? c) What is the amount of charge induced on the mica?

Short answer

Answer: a) The strength of the electric field in the mica is approximately \(1.28 \times 10^6\) V/m, b) The amount of free charge on the plates is approximately 6 nC, and c) The amount of charge induced on the mica is approximately 4.93 nC.

Step-by-step solution

Calculate the distance between the capacitor plates

To begin, we need to find the distance between the capacitor plates. We can use the formula for capacitance of a parallel plate capacitor: \(C = \epsilon_0 \epsilon_r \frac{A}{d}\), Where \(C\) is the capacitance, \(\epsilon_0\) is the vacuum permittivity (\(8.854 \times 10^{-12} \textrm{F/m}\)), \(\epsilon_r\) is the relative permittivity (dielectric constant) of the material between the plates, \(A\) is the plate area and \(d\) is the distance between plates. We are given \(C = 120\) pF, \(\epsilon_r = 5.4\), and \(A = 100\) cm\(^2\). Convert these values to SI units: \(C = 120 \times 10^{-12}\) F, \(A = 100 \times 10^{-4}\) m\(^2\). Rearrange the formula for \(d\): \(d = \epsilon_0 \epsilon_r \frac{A}{C}\) Plug in the values and calculate \(d\): \(d = (8.854 \times 10^{-12} \textrm{F/m})(5.4) \frac{100 \times 10^{-4} \mathrm{m}^2}{120\times 10^{-12} \mathrm{F}}\) \(d \approx 3.906 \times 10^{-5}\) m or 39.06 μm.

Calculate the electric field strength in the mica

Now we can calculate the electric field strength in the mica using the formula: \(E = \frac{V}{d}\), Where \(E\) is the electric field strength, \(V\) is the potential difference between the plates, and \(d\) is the distance between the plates. We are given the potential difference \(V = 50.0\) V and we found \(d = 3.906 \times 10^{-5}\) m in the previous step. Plug in these values to calculate the electric field strength: \(E = \frac{50.0 \mathrm{V}}{3.906 \times 10^{-5} \mathrm{m}}\) \(E \approx 1.28 \times 10^6\) V/m.

Calculate the free charge on the plates

Now we can find the amount of free charge on the plates. We can do this using the formula: \(Q = CV\), Where \(Q\) is the charge, \(C\) is the capacitance, and \(V\) is the potential difference between the plates. We are given \(C = 120 \times 10^{-12}\) F and \(V = 50\) V. Plug in these values to calculate the free charge on the plates: \(Q = (120 \times 10^{-12} \mathrm{F})(50 \mathrm{V})\) \(Q = 6 \times 10^{-9}\) C or 6 nC.

Calculate the charge induced on the mica

Finally, we can calculate the amount of charge induced on the mica. We can use the relationship between the dielectric constant, capacitance, and induced charge: \(Q_{induced} = Q(1 - \frac{1}{\epsilon_r})\) Where \(Q_{induced}\) is the induced charge, \(Q\) is the free charge on the plates, and \(\epsilon_r\) is the dielectric constant of the mica. Plug in the values we found in previous steps: \(Q_{induced} = 6 \times 10^{-9} \mathrm{C} (1 - \frac{1}{5.4})\) \(Q_{induced} \approx 4.928 \times 10^{-9}\) C or 4.93 nC. The answers to this exercise are: a) The strength of the electric field in the mica is approximately \(1.28 \times 10^6\) V/m, b) The amount of free charge on the plates is approximately 6 nC, and c) The amount of charge induced on the mica is approximately 4.93 nC.

Key concepts

Capacitance

Understanding the concept of capacitance is like getting to know the storage capacity of a water tank, but in the world of electronics, our 'water' is electric charge. So what is capacitance? Simply put, it's a measure of how much electric charge can be stored in a capacitor for a given voltage across its plates. Think of it as the electrical equivalent to the tank's volume.

With a parallel plate capacitor, such as the one described in our exercise with plates separated by a dielectric material (mica), capacitance is calculated using the formula:
\(C = \frac{\text{permittivity of the dielectric material} \times \text{plate area}}{\text{distance between the plates}}\).

In our case, the dielectric material is mica with a known dielectric constant, which, in simplistic terms, tells us how much better the mica stores electrical energy compared to empty space (a vacuum). It's an important factor because materials with higher dielectric constants can store more charge at the same voltage, enhancing the capacitor's capacity, much like building a taller tank to hold more water.

Dielectric Constant

The dielectric constant, also known as the relative permittivity, is another critical piece of our electrical storage puzzle. Think of the dielectric constant as a rating that tells us how good an insulator is compared to the benchmark performance of a vacuum. This rating also influences how strongly electric fields affect the material.

For instance, in our textbook problem, mica, a natural mineral, serves as the dielectric and has a dielectric constant of 5.40. This means that mica can store more than five times the amount of electrical energy that a vacuum can, given the same space between the plates. Crucially, this substantial increase in energy storage does not come with a corresponding increase in size or voltage, reflecting the efficiency offered by superior dielectrics like mica.

Mathematically, a higher dielectric constant amplifies the capacitance by the same factor when inserted within the capacitor. However, it's also important to note that while the dielectric constant boosts the energy storage, it doesn't impact the charge stored on the capacitor's plates directly—it requires the presence of an electric field.

Electric Field Strength

Finally, let's explore the electric field strength, a fundamental feature of our electrical landscape that is as basic and universal as gravity in the physical world. The electric field strength corresponds to the force that a charged particle would experience per unit charge in the presence of an electric field.

In terms of our parallel plate capacitor, the electric field strength in the dielectric (mica in this exercise) is determined by two things: the voltage applied across the plates and the distance separating them (

Formula

\(E = \frac{V}{d}\)). Simply put, if we increase the voltage, the electric field's 'push' on the charges gets stronger. Conversely, if we bring the plates closer together, we also ramp up the field strength — much like bringing two opposing magnets closer makes the attractive force between them far more noticeable.

In summary, the electric field strength tells us the potential energy per meter that a charge would experience in the field. In our solved exercise, we calculated an intense field strength of approximately 1.28 million volts per meter, exemplary of how potent electric fields can become within the confined space of a capacitor.