Question

A spherical capacitor is formed from two concentric spherical, conducting shells separated by vacuum. The inner sphere has radius 15.0 cm and the capacitance is 116 pF, (a) What is the radius of the outer sphere? (b) Pi the potential difference between the two spheres is 220 V, what is the magnitude of charge on each sphere?​

Answer 1

(a) The radius of the outer sphere is 45.0 cm.

(b) The magnitude of charge on each sphere is
`1.32 * 10^(^-^8^)` C.

Step-by-step explanation:

The capacitance (C) of a spherical capacitor is given by the formula
`\(C = \frac{{4\pi\epsilon r_1 r_2}}{{r_2 - r_1}}\), where \(r_1\) and \(r_2\)` are the radii of the inner and outer spheres, respectively, and
`\(\epsilon\)` is the permittivity of the medium between the spheres.

In this problem, we are given that
`\(r_1 = 15.0 \ cm\)` and
`\(C = 116 \ pF\)`. First, convert the capacitance to farads by multiplying it by
`\(10^(-12)\): \(C = 116 * 10^(-12) \ F\)`.

Now, rearrange the formula to solve for \(r_2\): \(r_2 = \frac{{C \cdot (r_1 + \sqrt{{C \cdot \frac{1}{{4\pi\epsilon}}}})}}{{C \cdot \sqrt{{\frac{1}{{4\pi\epsilon}}}}}\).

Substitute the known values into the formula: \(r_2 = \frac{{15.0 + \sqrt{{116 \cdot 10^{-12} \cdot \frac{1}{{4\pi\epsilon}}}}}}{{\sqrt{{\frac{1}{{4\pi\epsilon}}}}}\).

After calculating, the radius of the outer sphere (\(r_2\)) is found to be \(45.0 \ cm\).

To determine the charge on each sphere (\(Q\)), we use the formula \(Q = C \cdot V\), where \(V\) is the potential difference between the spheres. Given \(V = 220 \ V\), calculate \(Q\): \(Q = 116 \times 10^{-12} \ F \times 220 \ V\).

The magnitude of the charge on each sphere is found to be \(1.32 \times 10^{-8} \ C\).

In summary, the radius of the outer sphere is \(45.0 \ cm\) and the magnitude of the charge on each sphere is \(1.32 \times 10^{-8} \ C\).

Answer 2

The magnitude of charge
`\( Q \)` on each sphere, with a potential difference of
`\( 220 \)` volts, is approximately
`\( 25.52 \)` nanoCoulombs (nC).

To solve this problem, let's break it down into the two parts you've outlined:

Part (a): Finding the Radius of the Outer Sphere

Step 1: Understand the Formula for Capacitance of a Spherical Capacitor

The capacitance
`\( C \)` of a spherical capacitor with an inner radius
`\( r_a \)` and an outer radius
`\( r_b \)` in a vacuum is given by:

`\[ C = 4\pi \varepsilon_0 (r_a r_b)/(r_b - r_a) \]`

`where \( \varepsilon_0 \) is the permittivity of free space (\( \varepsilon_0 \approx 8.85 * 10^(-12) \, \text{F/m} \)).`

Step 2: Plug in Known Values and Solve for
`\( r_b \)`

Given:
`\( C = 116 \, \text{pF} = 116 * 10^(-12) \, \text{F} \), \( r_a = 15.0 \, \text{cm} = 0.15 \, \text{m} \).`

Rearrange the formula to solve for
`\( r_b \)` :

`\[ r_b = (C)/(4\pi \varepsilon_0 r_a) + r_a \]`

Part (b): Finding the Magnitude of Charge on Each Sphere

Step 1: Understand the Relationship Between Capacitance, Voltage, and Charge

The capacitance
`\( C \) i`s related to the potential difference
`\( V \)` and the charge
`\( Q \)` by:

`\[ C = (Q)/(V) \]`

Rearrange to solve for
`\( Q \):`

`\[ Q = C \cdot V \]`

Step 2: Plug in Known Values to Find
`\( Q \)`

Given:
`\( V = 220 \, \text{V} \), and \( C = 116 \, \text{pF} \) (from part (a)).`

Calculate
`\( Q \):`

`\[ Q = 116 * 10^(-12) \, \text{F} * 220 \, \text{V} \]`

Let's perform these calculations.

Solution

Part (a): Radius of the Outer Sphere

The radius of the outer sphere
`\( r_b \)` is approximately
`\( 7.10 \)` meters.

Part (b): Magnitude of Charge on Each Sphere

The magnitude of charge
`\( Q \)` on each sphere, with a potential difference of
`\( 220 \)` volts, is approximately
`\( 25.52 \)` nanoCoulombs (nC).