Question

Two metal spheres have charges of equal magnitude, $3.2 \times 10^{-14} \mathrm{C},$ but opposite sign. If the potential difference between the two spheres is \(4.0 \mathrm{mV},\) what is the capacitance? [Hint: The "plates" are not parallel, but the definition of capacitance holds.]

Short answer

Answer: The capacitance between the two metal spheres is \(8 \times 10^{-12} \mathrm{F}\).

Step-by-step solution

Write down the definition of capacitance

The definition of capacitance is given by the formula: \(C = \frac{Q}{V}\) where C is the capacitance, Q is the charge, and V is the potential difference.

Convert the potential difference to volts

The potential difference between the spheres is given as \(4.0 \mathrm{mV}\). To use it in our calculation, we need to convert it to volts. 1 mV = \(10^{-3}\) V So, \(4.0 \mathrm{mV} = 4.0 \times 10^{-3} \mathrm{V}\)

Apply the known values

We are given the charge on the spheres which is \(3.2 \times 10^{-14} \mathrm{C}\) and the potential difference converted to volts which is \(4.0 \times 10^{-3} \mathrm{V}\). Substitute the values into the capacitance formula: \(C = \frac{3.2 \times 10^{-14} \mathrm{C}}{4.0 \times 10^{-3} \mathrm{V}}\)

Calculate the capacitance

Now, perform the division to find the value of the capacitance: \(C = \frac{3.2 \times 10^{-14}}{4.0 \times 10^{-3}} = 8 \times 10^{-12} \mathrm{F}\) The capacitance of the two metal spheres is \(8 \times 10^{-12} \mathrm{F}\).