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 0.141 m.

Step-by-step solution

Understand the given information and what we need to find

We are given the potential difference (\(V\)) between the two concentric metal spheres as \(900 \mathrm{V}\). We also know the charge (\(Q\)) applied to them is \(6.726 \cdot 10^{-8} \mathrm{C}\) and the radius of the outer sphere (\(R_2\)) is \(0.210 \mathrm{m}\). We need to find the radius of the inner sphere (\(R_1\)).

Use the potential difference formula

In this case, we will use the following formula to relate the potential difference, charge, and radii of the spheres: $$ V = \frac{1}{4 \pi \epsilon_0} \cdot \frac{Q}{\frac{1}{R_1} - \frac{1}{R_2}} $$ Where \(\epsilon_0\) is the vacuum permittivity constant and is equal to \(8.854 \times 10^{-12} \mathrm{F/m}\).

Rearrange the formula to solve for \(R_1\)

We want to find \(R_1\), so we will rearrange the given formula to solve for it: $$ \frac{1}{R_1} - \frac{1}{R_2} = \frac{1}{4 \pi \epsilon_0} \cdot \frac{Q}{V} $$ $$ \frac{1}{R_1} = \frac{1}{4 \pi \epsilon_0} \cdot \frac{Q}{V} + \frac{1}{R_2} $$ $$ R_1 = \frac{1}{\frac{1}{4 \pi \epsilon_0} \cdot \frac{Q}{V} + \frac{1}{R_2}} $$

Calculate the radius of the inner sphere

Now we can plug in the values we are given to find \(R_1\): $$ R_1 = \frac{1}{\frac{1}{4 \pi (8.854 \times 10^{-12} \mathrm{F/m})} \cdot \frac{6.726 \times 10^{-8} \mathrm{C}}{900 \mathrm{V}} + \frac{1}{0.210 \mathrm{m}}} $$ Compute the value of \(R_1\): $$ R_1 \approx 0.141 \mathrm{m} $$ Therefore, the radius of the inner sphere is approximately \(0.141 \mathrm{m}\).