Question

Two concentric metal spheres are found to have a potential difference of \(900 . \mathrm{V}\) when a charge of \(6.726 \cdot 10^{-8} \mathrm{C}\) is applied to them. The radius of the outer sphere is \(0.210 \mathrm{~m}\). What is the radius of the inner sphere?

Short answer

Answer: The radius of the inner sphere is approximately 0.0994 meters.

Step-by-step solution

Use the capacitance formula for two spherical conductors

The capacitance of a spherical capacitor is given by the formula: $$ C = 4\pi \epsilon_0 \frac{r_1r_2}{r_2-r_1} $$ Where \(C\) is the capacitance, \(\epsilon_0\) is the permittivity of free space, \(r_1\) and \(r_2\) are the radii of the inner and outer spheres, respectively.

Use the relationship between voltage, capacitance, and charge

We have the relationship: $$ V = \frac{Q}{C} $$ Where \(V\) is the potential difference, \(Q\) is the charge, and \(C\) is the capacitance.

Rewrite the capacitance equation for the given variables

Using the given data and given equations, \(Q = 6.726 \cdot 10^{-8}\;\mathrm{C}\), \(r_2 = 0.210\;\mathrm{m}\), \(V = 900\;\mathrm{V}\), and the value for \(\epsilon_0 = 8.854 \times 10^{-12}\;\mathrm{F/m}\), we can rewrite the capacitance formula as follows: $$ C = 4\pi(8.854 \times 10^{-12}\;\mathrm{F/m}) \frac{r_1(0.210\;\mathrm{m})}{0.210\;\mathrm{m} - r_1} $$

Substitute the given values and solve for \(r_1\)

Now let's substitute the given values and use the relationship between voltage, capacitance, and charge to find the radius of the inner sphere, \(r_1\): $$900\;\mathrm{V} = \frac{6.726 \cdot 10^{-8}\;\mathrm{C}}{4\pi(8.854 \times 10^{-12}\;\mathrm{F/m}) \frac{r_1(0.210\;\mathrm{m})}{0.210\;\mathrm{m} - r_1}}$$ Solve the equation for \(r_1\), we get: $$ r_1 = \frac{0.210\;\mathrm{m}}{1+\frac{6.726 \cdot 10^{-8}\;\mathrm{C}}{4\pi(8.854 \times 10^{-12}\;\mathrm{F/m})(900\;\mathrm{V})}} \approx 0.0994\;\mathrm{m} $$

Final Answer

The radius of the inner sphere is approximately \(0.0994\;\mathrm{m}\).

Key concepts

Capacitance of Spherical Conductors

Understanding the capacitance of spherical conductors is crucial for grasping complex concepts in electromagnetism. The capacitance is a measure of how much electric charge a conductor can hold at a certain voltage.

In the case of spherical conductors, we have two metal spheres, one inside the other, with air or some other dielectric material between them. The key formula to remember is: $$ C = 4\text{π}ε_0 \frac{r_1r_2}{r_2 - r_1} $$where \(C\) is capacitance, \(ε_0\) is the permittivity of free space, and \(r_1\) and \(r_2\) are the radii of the inner and outer spheres, respectively.

To simplify, consider the spheres like a storage device for electric charge—the larger the spheres and the closer they are without touching, the more charge they can store at a lower voltage, leading to higher capacitance. This concept is directly connected to how capacitors are used in electronic circuits to store and release energy.

Permittivity of Free Space

The permittivity of free space, often symbolized as \(ε_0\), is a constant that plays a fundamental role in the calculation of capacitance. It represents the ability of the vacuum to allow electric field lines to flow through it.

The value of \(ε_0\) is about \(8.854 \times 10^{-12} F/m\) (farads per meter), where a farad is the unit of capacitance. When dealing with capacitors, think of permittivity as a factor that influences how much energy the electric field can hold. It's like the 'thickness' of the 'electric field sponge' that sits in between the conducting plates.

In our textbook example with spherical conductors, \(ε_0\) appears in the capacitance formula, indicating that space itself, even in the absence of any material, affects how electric charges interact at a distance. This characteristic is essential for understanding electric fields and how capacitors work.

Potential Difference and Charge Relationship

The potential difference (voltage) and charge relationship in a capacitor is a critical concept in circuit analysis. This relationship ensures we understand how a change in one quantity affects another.

Expressed mathematically, the relationship is given by $$ V = \frac{Q}{C} $$where \(V\) is the potential difference in volts, \(Q\) is the charge in coulombs, and \(C\) is the capacitance in farads. Here's a way to visualize it: voltage acts as the 'pressure' that pushes charges onto the capacitor, and capacitance is 'how much space' there is to store those charges.

When solving problems, such as finding the radius of a spherical capacitor's inner sphere, we manipulate this relationship, substituting values for charge and capacitance. We can then see how an electric potential can 'hold' a certain amount of charge based on the capacitance, which, in our exercise, is defined by the spheres' radii and the material between them. It's akin to adjusting the size of a water balloon (the capacitor) to see how much water (charge) it can hold at a particular level of stretching (voltage).