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An air-filled spherical capacitor is constructed with inner and outer shell radii of 7.00 and \(14.0 \mathrm{cm},\) respectively. (a) Calculate the capacitance of the device. (b) What potential difference between the spheres results in a charge of 4.00\(\mu \mathrm{C}\) on the capacitor?

Short Answer

Expert verified
The capacitance 'C' is calculated using the spherical capacitor formula. The potential difference 'V' is found using the charge-capacitance relationship with the capacitance from the previous step and given charge.

Step by step solution

01

Understanding the problem

We are given an air-filled spherical capacitor with inner radius 'a' of 7.00 cm and outer radius 'b' of 14.0 cm. We have to find (a) the capacitance 'C' of the capacitor, and (b) the potential difference 'V' between the spheres given a charge 'Q' of 4.00 µC.
02

Calculating the capacitance

The capacitance 'C' of a spherical capacitor is given by the formula \(C = 4\pi\epsilon_0\frac{ab}{b-a}\), where \(\epsilon_0\) is the vacuum permittivity with a value of \(8.854 \times 10^{-12} F/m\). Substitute the values of 'a' and 'b' given in the problem into the formula to calculate 'C'.
03

Substitute and calculate capacitance

First, convert the radii into meters: 'a' = 7.00 cm = 0.0700 m and 'b' = 14.0 cm = 0.140 m. Now we can substitute the values to find 'C': \(C = 4\pi(8.854 \times 10^{-12})\frac{0.0700 \times 0.140}{0.140 - 0.0700}\), then calculate the result.
04

Solving for the potential difference

The relationship between charge 'Q', capacitance 'C', and potential difference 'V' is given by \(Q = C \times V\). To find 'V', rearrange the equation to \(V = \frac{Q}{C}\) and substitute the value of the charge 'Q' (in Coulombs) and the capacitance 'C' (from Step 3) into the formula to calculate 'V'.
05

Calculate the potential difference

First convert the charge into Coulombs: 4.00 µC = 4.00 × 10^{-6} C. Now, use the capacitance calculated in Step 3: \(V = \frac{4.00 \times 10^{-6} C}{C calculated in Step 3}\) to find the potential difference 'V'.

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Key Concepts

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

Capacitance Calculation
Understanding the capacitance calculation for a spherical capacitor involves some essential physical principles. A spherical capacitor comprises two concentric spherical conducting shells separated by an insulating material—in this case, air.

The formula for calculating the capacitance 'C' is \( C = 4\pi\epsilon_0\frac{ab}{b-a} \), where \( \epsilon_0 \) (commonly referred to as the vacuum permittivity) is a constant with the approximate value of \( 8.854 \times 10^{-12} \text{F/m} \), 'a' represents the radius of the inner shell, and 'b' is the radius of the outer shell. This equation is a direct consequence of the Gauss's law, considering the spherical symmetry of the electric field between the conductors.

To apply our understanding, let's take our example where the inner radius 'a' is 0.0700 meters and the outer radius 'b' is 0.140 meters. The capacitance can be calculated by inserting these values into our formula. Short and neatly explained, this shows how dimensions of a spherical capacitor directly affect its capacitance.
Potential Difference
The potential difference, often denoted by 'V', measures the amount of electric potential energy per unit charge between two points in an electric field. In the context of our spherical capacitor, the potential difference between the two spherical shells indicates the energy required to move a unit charge from one shell to another.

According to the basic relationship \( Q = C \times V \), where 'Q' is the electric charge, 'C' is the capacitance, and 'V' is the potential difference, we can rearrange the formula to solve for 'V': \( V = \frac{Q}{C} \). In practical terms, if we know the charge deposited on the capacitor and its capacitance, we can determine the potential difference. For example, with a charge of 4.00 µC (or \( 4.00 \times 10^{-6} C \) in Coulombs) already calculated and the capacitance from our previous step, we can compute the precise electric potential that builds up between the shells of the spherical capacitor.
Electric Charge

Quantization and Conservation of Charge

Electric charge 'Q' is a fundamental property of matter determining how it interacts within an electric field. Charge is 'quantized,' meaning it exists in discrete packets (multiples of the elementary charge \( e \) ). On a macroscopic scale, charges are usually measured in Coulombs.

Charge on Capacitors

In the context of capacitors, charge refers to the amount of electric charge stored on the plates due to the presence of an electric potential. For spherical capacitors, charges on the inner and outer shells are equal in magnitude but opposite in sign, following the conservation of charge principle, which states that the total electric charge in an isolated system remains constant.

By applying the known values of capacitance and potential difference, we can solve for the electric charge using the rearranged equation \( Q = C \times V \). This equation provides a direct link to the energy stored in an electrostatic field of the capacitor, underlying the concepts of electrostatic energy storage and power management in electronic circuits.

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Most popular questions from this chapter

An isolated charged conducting sphere of radius 12.0 \(\mathrm{cm}\) creates an electric field of \(4.90 \times 10^{4} \mathrm{N} / \mathrm{C}\) at a distance 21.0 \(\mathrm{cm}\) from its center. (a) What is its surface charge density? (b) What is its capacitance?

An isolated capacitor of unknown capacitance has been charged to a potential difference of 100 \(\mathrm{V}\) . When the charged capacitor is then connected in parallel to an uncharged \(10.0-\mu \mathrm{F}\) capacitor, the potential difference across the combination is 30.0 \(\mathrm{V}\) . Calculate the unknown capacitance.

Two conductors having net charges of \(+10.0 \mu \mathrm{C}\) and \(-10.0 \mu \mathrm{C}\) have a potential difference of 10.0 \(\mathrm{V}\) between them. (a) Determine the capacitance of the system. (b) What is the potential difference between the two conductors if the charges on each are increased to \(+100 \mu \mathrm{C}\) and \(-100 \mu \mathrm{C}\) ?

Capacitors \(C_{1}=6.00 \mu \mathrm{F}\) and \(C_{2}=2.00 \mu \mathrm{F}\) are charged as a parallel combination across a \(250-\mathrm{V}\) battery. The capacitors are disconnected from the battery and from each other. They are then connected positive plate to negative plate and negative plate to positive plate. Calculate the resulting charge on each capacitor.

A parallel-plate capacitor of plate separation \(d\) is charged to a potential difference \(\Delta V_{0} .\) A dielectric slab of thickness \(d\) and diectric constant \(\kappa\) is introduced between the plates while the battery remains connected to the plates. (a) Show that the ratio of energy stored after the dielectric is introduced to the energy stored in the empty capacitor is \(U / U_{0}=\kappa\) Give a physical explanation for this increase in stored energy. (b) What happens to the charge on the capacitor? (Note that this situation is not the same as in Example 26.7 , in which the battery was removed from the circuit before the dielectric was introduced.)

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