Question

A dielectric slab with thickness \(d\) and dielectric constant \(\kappa=2.31\) is inserted in a parallel place capacitor that has been charged by a \(110 .-\mathrm{V}\) battery and having area \(A=\) \(100, \mathrm{~cm}^{2},\) and separation distance \(d=2.50 \mathrm{~cm}\). a) Find the capacitance, \(C,\) the potential difference, \(V,\) the electric field, \(E,\) the total charge stored on the capacitor \(Q\), and electric potential energy stored in the capacitor, \(U\), before the dielectric material is inserted. b) Find \(C, V, E, Q,\) and \(U\) when the dielectric slab has been inserted and the battery is still connected. c) Find \(C, V, E, Q\) and \(U\) when the dielectric slab is in place and the battery is disconnected.

Short answer

After inserting the dielectric slab and keeping the battery connected, calculate the same values again. Finally, calculate the values after inserting the dielectric slab and disconnecting the battery. Answer: Before the dielectric is inserted: - Capacitance (C) = 3.54 x 10^-12 F - Potential difference (V) = 110 V - Electric field (E) = 4400 V/m - Total charge stored (Q) = 3.89 x 10^-10 C - Electrical potential energy (U) = 2.14 x 10^-8 J After inserting the dielectric slab and keeping the battery connected: - Capacitance (C') = 8.17 x 10^-12 F - Potential difference (V') = 110 V - Electric field (E') = 1904.33 V/m - Total charge stored (Q') = 8.99 x 10^-10 C - Electrical potential energy (U') = 4.93 x 10^-8 J After inserting the dielectric slab and disconnecting the battery: - Capacitance (C'') = 8.17 x 10^-12 F - Potential difference (V'') = 110 V - Electric field (E'') = 825.25 V/m - Total charge stored (Q'') = 8.99 x 10^-10 C - Electrical potential energy (U'') = 4.93 x 10^-8 J

Step-by-step solution

Calculate the capacitance (C) without dielectric

We can use the formula for the capacitance of a parallel plate capacitor without dielectric: \(C=\frac{\epsilon_0A}{d}\) Where \(\epsilon_0=8.85\times10^{-12}\ \mathrm{F/m^2}\) is the vacuum permittivity, \(A=100\ \mathrm{cm}^2=0.01\ \mathrm{m}^2\) is the area, and \(d=2.50\ \mathrm{cm}=0.025\ \mathrm{m}\) is the separation distance. So, \(C= \frac{8.85\times10^{-12} \times 0.01}{0.025} = 3.54\times10^{-12}\ \mathrm{F}\).

Calculate the potential difference (V)

The potential difference is given as: \(V=110\ \mathrm{V}\).

Calculate the electric field (E)

Using the formula for the electric field in a capacitor without dielectric, we have: \(E=\frac{V}{d}=\frac{110}{0.025} = 4400\ \mathrm{V/m}\).

Calculate the total charge stored (Q)

Now, we can calculate the total charge stored in the capacitor using the formula: \(Q=CV = 3.54\times10^{-12} \times 110 = 3.89\times10^{-10}\ \mathrm{C}\).

Calculate the electrical potential energy stored (U)

Finally, we can calculate the electrical potential energy using the formula: \(U=\frac{1}{2}CV^2 =\frac{1}{2} \times 3.54\times10^{-12} \times (110)^2 = 2.14\times10^{-8}\ \mathrm{J}\). # Part b: After the dielectric slab is inserted and the battery is still connected #

Calculate the capacitance (C') with dielectric

Using the capacitance with dielectric formula, we have: \(C'=\kappa C=2.31 \times 3.54\times10^{-12} =8.17\times10^{-12}\ \mathrm{F}\).

Calculate the potential difference (V')

Since the battery is still connected, the potential difference remains the same: \(V'=V=110\ \mathrm{V}\).

Calculate the electric field (E')

By calculating the electric field with dielectric, we get: \(E'=\frac{E}{\kappa}=\frac{4400}{2.31} =1904.33\ \mathrm{V/m}\).

Calculate the total charge stored (Q')

Calculate the total charge stored in the capacitor: \(Q'=C'V'=8.17\times10^{-12}\times110 = 8.99\times10^{-10}\ \mathrm{C}\).

Calculate the electrical potential energy (U')

Calculate the electrical potential energy stored in the capacitor: \(U'=\frac{1}{2}C'(V')^2=\frac{1}{2} \times 8.17\times10^{-12} \times (110)^2=4.93\times10^{-8}\ \mathrm{J}\). # Part c: After the dielectric slab is inserted and the battery is disconnected #

Capacitance remains the same (C')

The capacitance won't change since we still have the dielectric inserted: \(C''=C'= 8.17\times10^{-12}\ \mathrm{F}\).

Calculate the total charge stored (Q'')

Since the battery is disconnected, the total charge remains the same: \(Q''=Q'= 8.99\times10^{-10}\ \mathrm{C}\).

Calculate the potential difference (V'')

Use the formula for total charge stored in a capacitor: \(V''=\frac{Q''}{C''}=\frac{8.99\times10^{-10}}{8.17\times10^{-12}}= 110\ \mathrm{V}\).

Calculate the electric field (E'')

Calculate the electric field using the formula for electric field with dielectric: \(E''=\frac{E'}{\kappa}=\frac{1904.33}{2.31} = 825.25\ \mathrm{V/m}\).

Calculate the electrical potential energy (U'')

Calculate the electrical potential energy stored in the capacitor: \(U''=\frac{1}{2}C''(V'')^2=\frac{1}{2} \times 8.17\times10^{-12} \times (110)^2= 4.93\times10^{-8}\ \mathrm{J}\).

Key concepts

Capacitance Calculation

When dealing with a parallel plate capacitor, the capacitance is a measure of how much electric charge the capacitor can store per unit of electrical potential difference between its plates. The basic formula for a capacitor without a dielectric material is given by:
\[C = \frac{\epsilon_0 A}{d}\] where \(\epsilon_0\) represents the permittivity of free space, \(A\) is the plate area, and \(d\) is the separation distance between the plates.

When a dielectric slab with dielectric constant \(\kappa\) is introduced between the plates, the capacitance increases and is calculated by the modified formula: \[C' = \kappa C\]
This illustrates how dielectric materials influence the ability of a capacitor to store charge, by reducing the electric field within and thus allowing for more charge to be stored for the same applied voltage.

Electric Field in Capacitors

The electric field within a capacitor is a measure of the electric force per unit charge between the plates. For a capacitor without a dielectric, the electric field is simply the voltage divided by the separation distance: \[E = \frac{V}{d}\]

However, inserting a dielectric material affects the electric field by a factor equal to the inverse of the dielectric constant. With the dielectric present, the electric field is reduced to: \[E' = \frac{E}{\kappa}\]
This reduced electric field has implications for the energy stored in the capacitor and how it interacts with electric charges. Incorporating a dielectric therefore serves as an effective method to enhance the capacitor's properties without modifying its physical size.

Electrical Potential Energy

The electrical potential energy, denoted by \(U\), is the energy stored in a capacitor due to the separation of charges on its plates. This is inherently related to the work done in charging the capacitor. For a simple parallel plate capacitor, the energy can be calculated using the formula: \[U = \frac{1}{2}CV^2\]

After inserting a dielectric slab, while keeping the battery connected, the capacitance increases which in turn increases the potential energy stored in the capacitor. Therefore, we observe an interesting phenomenon - the insertion of a dielectric in a charging capacitor does not alter the voltage but boosts the energy stored, rendering the formula: \[U' = \frac{1}{2}C'V^2\]
Highlighting this principle is crucial for students to grasp how dielectrics enhance the storage capacity of capacitors.

Charge Stored in Capacitor

The total charge stored in a capacitor, represented by \(Q\), is directly proportional to both the capacitance and the voltage applied across the capacitor. The relationship is quantified by the equation: \[Q = CV\]

When a capacitor with a dielectric connected to a battery is charged, the dielectric increases the capacitance, allowing the capacitor to store more charge without changing the voltage. This characteristic is reflected in our calculation by: \[Q' = C'V\]
When the battery is disconnected, the charge on the capacitor remains constant regardless of whether the dielectric constant changes, since no additional charge can flow into the system. Understanding this concept elucidates how and why capacitors are essential components in circuits that store and release energy.