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A parallel-plate capacitor has capacitance \(C=12.5 \mathrm{pF}\) when the volume between the plates is filled with air. The plates are circular, with radius \(3.00 \mathrm{~cm}\). The capacitor is connected to a battery, and a charge of magnitude \(25.0 \mathrm{pC}\) goes onto each plate. With the capacitor still connected to the battery, a slab of dielectric is inserted between the plates, completely filling the space between the plates. After the dielectric has been inserted, the charge on each plate has magnitude \(45.0 \mathrm{pC}\). (a) What is the dielectric constant \(K\) of the dielectric? (b) What is the potential difference between the plates before and after the dielectric has been inserted? (c) What is the electric field at a point midway between the plates before and after the dielectric has been inserted?

Short Answer

Expert verified
The dielectric constant is 1.8. The potential difference before the dielectric is inserted is 2.0 V and after the dielectric is inserted is 2.4 V. The electric field before the dielectric is inserted is 0.67 V/cm and after the dielectric is inserted is 0.8 V/cm.

Step by step solution

01

Determine the dielectric constant

Capacitance of a capacitor when filled with dielectric becomes \(C' = KC\), where K is the dielectric constant. Also, the charge \(Q' = C'V = C'Q/C = KCQ/C = KQ\) after the introduction of the dielectric. Therefore, by comparing the initial and final charge \(K = Q'/Q = 45.0 pC / 25.0 pC = 1.8 \)
02

Determine the potential difference before and after the dielectric is inserted

Potential difference across the capacitor can be calculated by \(V = Q/C\) and \(V' = Q'/C' = Q'/(KC) = Q'/KQ = 1/K\). From these formulas, we can calculate the potential difference before the dielectric to be \(V = 25.0 pC / 12.5 pF = 2.0 V\), and after the dielectric to be \(V' = 45.0 pC / (1.8 * 12.5 pF) = 2.4 V\)
03

Determine the electric field before and after the dielectric

Electric field can be calculated by \(E = V/d\) and \(E' = V'/d\), where \(d\) is the distance between the plates. For a parallel plate capacitor, \(d = radius = 3.0 cm\). Before the dielectric is inserted, \(E = 2.0 V / 3.0 cm = 0.67 V/cm\), and after the dielectric is inserted, \(E' = 2.4 V / 3.0 cm = 0.8 V/cm\)

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Dielectric Constant
In conjunction with capacitors, a dielectric plays a pivotal role in determining how much electrical energy can be stored. The dielectric constant (\( K \text{ or } \.epsilon_r \text{ (relative permittivity)} \) ) is a measure that reflects how much a dielectric material can reduce the electric field within a capacitor compared to the electric field in a vacuum. It's a dimensionless number that indicates the increase in capacitance that a material provides when it fills the space between the capacitor plates.

Imagine inserting a non-conductive substance between the plates of a charged capacitor; the dielectric reduces the electric field and allows the capacitor to store more charge at the same voltage. This is because the dielectric's molecules align in a way that reduces the effective field within the material, promoting increased charge accumulation on the plates for the same applied voltage. The higher the dielectric constant of a material, the better the material is at insulating the electric field, which, in turn, leads to greater energy storage potential in the capacitor.
Electric Field
The electric field between the plates of a parallel-plate capacitor can be envisioned as a uniform force field that interacts with charges within the space. This field is created by the separation of charges on the plates and exerts a force on any other charges that enter between them. Mathematically, the strength of the electric field (\( E \) ) is defined as the force per unit charge and in the context of a parallel-plate capacitor is given simply by the equation \( E = V/d \) where \( V \) represents the potential difference across the plates, and \( d \) is the distance between them.

When a dielectric material is inserted in the capacitor, it affects the electric field. The field inside the dielectric is decreased, which conceptually can be understood as the dielectric 'weakening' the field due to its molecular structure. This process allows for the same voltage to maintain a stronger force on the stored charges, leading to a higher capacitance.
Potential Difference
The term potential difference, often referred to as voltage (\( V \) ), is a measurable quantity that describes the energy difference per unit charge between two points in an electric field. In a capacitor, it's the electric potential energy difference between the two plates and indicates how much work would be done to move a charge between them.

When a capacitor is connected to a battery, the potential difference between the plates is set by the battery's voltage. Inserting a dielectric changes the capacitor's ability to store charge without altering this imposed potential difference immediately. However, once the dielectric is fully inserted and the capacitor reaches a new equilibrium, the potential difference can change if the capacitor is disconnected from the battery. The capacitor with the dielectric can hold more charge for the same potential difference, or equivalently, achieve the same charge at a lower potential difference, demonstrating the dielectric's impact on the capacitor's properties.
Capacitance
The concept of capacitance encompasses a capacitor's capability to hold an electric charge. Quantified with the unit Farad (\( F \) ), capacitance is given by the equation \( C = Q/V \) where \( Q \) is the charge stored, and \( V \) is the potential difference across the capacitor. For a parallel-plate capacitor with no dielectric, its capacitance is directly proportional to the plate area and inversely proportional to the distance between the plates.

Introducing a dielectric into the capacitor increases its capacitance, allowing it to store more charge at the same voltage. The equation \( C' = KC \) signifies the new capacitance (\( C' \) ) after introducing the dielectric, with \( C \) representing the original capacitance in the absence of a dielectric. The dielectric constant (\( K \) ) is imperative since it quantifies how much the capacitance has increased due to the presence of the dielectric material.

Understanding these fundamental principles of a capacitor's behaviour when interacting with dielectrics affords valuable insights for students diving into the workings of electronic circuits and the storage of electrical energy.

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Most popular questions from this chapter

DATA Your electronics company has several identical capacitors with capacitance \(C_{1}\) and several others with capacitance \(C_{2}\). You must determine the values of \(C_{1}\) and \(C_{2}\) but don't have access to \(C_{1}\) and \(C_{2}\) individually. Instead, you have a network with \(C_{1}\) and \(C_{2}\) connected in series and a network with \(C_{1}\) and \(C_{2}\) connected in parallel. You have a \(200.0 \mathrm{~V}\) battery and instrumentation that measures the total \(\mathrm{en}\) ergy supplied by the battery when it is connected to the network. When the parallel combination is connected to the battery, \(0.180 \mathrm{~J}\) of energyis stored in the network. When the series combination is connected. \(0.0400 \mathrm{~J}\) of energy is stored. You are told that \(C_{1}\) is greater than \(C_{2}\) (a) Calculate \(C_{1}\) and \(C_{2}\). (b) For the series combination, does \(C_{1}\) or \(C_{2}\) store more charge, or are the values equal? Does \(C_{1}\) or \(C_{2}\) store more energy, or are the values equal? (c) Repeat part (b) for the parallel combination.

A \(5.00 \mathrm{pF},\) parallel-plate, air-filled capacitor with circular to a \(12.0 \mathrm{~V}\) battery. (a) What is the charge on each plate? (b) How much charge would be on the plates if their separation were doubled while the capacitor remained connected to the battery? (c) How much charge would be on the plates if the capacitor were connected to the \(12.0 \mathrm{~V}\) battery after the radius of each plate was doubled without changing their separation?

Capacitance of a Thundercloud. The charge center of a thundercloud, drifting \(3.0 \mathrm{~km}\) above the earth's surface, contains \(20 \mathrm{C}\) of negative charge. Assuming the charge center has a radius of \(1.0 \mathrm{~km}\), and modeling the charge center and the earth's surface as parallel plates, calculate: (a) the capacitance of the system; (b) the potential difference between charge center and ground; (c) the average strength of the electric field between cloud and ground; (d) the electrical energy stored in the system.

A fuel gauge uses a capacitor to determine the height of the fuel in a tank. The effective dielectric con- stant \(K_{\text {eff }}\) changes from a value of 1 when the tank is empty to a value of \(K\), the dielectric constant of the fuel, when the tank is full. The appropriate electronic circuitry can determine the effective dielectric constant of the combined air and fuel between the capacitor plates. Each of the two rectangular plates has a width \(w\) and a length \(L\) (Fig. \(\mathbf{P 2 4 . 6 6}\) ). The height of the fuel between the plates is \(h\). You can ignore any fringing effects. (a) Derive an expression for \(K_{\text {eff }}\) as a function of \(h\). (b) What is the effective dielectric constant for a tank \(\frac{1}{4}\) full, \(\frac{1}{2}\) full, and \(\frac{3}{4}\) full if the fuel is gasoline \((K=1.95) ?\) (c) Repeat part (b) for methanol \((K=33.0)\). (d) For which fuel is this fuel gauge more practical?

CALC Two conducting plates with area \(A\) are separated by distance \(d\). Between the plates is a material with a dielectric constant that varies linearly from a value of unity next to one plate to \(K\) next to the other plate. (a) What is the capacitance of this device? [Hint: We can envision this as a continuum of capacitors with differential plate separation connected in series. The reciprocal of the capacitance of a differential slice is then \(d x /\left(K(x) \epsilon_{0} A\right),\) where \(K(x)\) is the dielectric constant specific to that locale. \(]\) (b) Show that this result matches the expected result when \(K \rightarrow 1\)

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