Question

A certain capacitor stores \(450 \mathrm{J}\) of energy when it holds $8.0 \times 10^{-2} \mathrm{C}$ of charge. What is (a) the capacitance of this capacitor and (b) the potential difference across the plates?

Short answer

Answer: The capacitance of the capacitor is approximately 4.0 x 10^(-5) F, and the potential difference across the plates is 2000 V.

Step-by-step solution

Find the capacitance using the energy stored formula

First, let's rewrite the energy stored formula to solve for capacitance: \(C = \frac{2W}{V^2}\) We know the energy stored (W) is \(450 J\), but we don't have the potential difference (V) yet. To find V, we will use the definition of capacitance formula by rearranging it to solve for V: \(V = \frac{Q}{C}\) However, we don't have C yet. We will rearrange the energy stored formula again to solve for potential difference: \(V = \sqrt{\frac{2W}{C}}\) Now, we can rewrite the definition of capacitance formula with this new expression for V: \(C = \frac{Q}{\sqrt{\frac{2W}{C}}}\) We can now solve for capacitance C by substituting the given values of energy stored (W) and charge (Q): \(C = \frac{8.0 \times 10^{-2} \mathrm{C}}{\sqrt{\frac{2 \times 450 \mathrm{J}}{C}}}\) By squaring both sides of the equation, we obtain: \(C^2 = \frac{(8.0 \times 10^{-2} \mathrm{C})^2}{\frac{2 \times 450 \mathrm{J}}{C}}\) Now we can cross-multiply and solve for C: \(C^3 = (8.0 \times 10^{-2} \mathrm{C})^2 \times \frac{1}{2 \times 450 \mathrm{J}}\) \(C = \sqrt[3]{\frac{(8.0 \times 10^{-2} \mathrm{C})^2}{2 \times 450 \mathrm{J}}} = 3.999 \times 10^{-5} \mathrm{F}\) (approximately) So the capacitance of the capacitor is approximately \(4.0 \times 10^{-5} \mathrm{F}\).

Find the potential difference using the definition of capacitance formula

Now, we can use the definition of capacitance formula we rearranged earlier to solve for the potential difference: \(V = \frac{Q}{C}\) We know both the charge (Q) and the calculated capacitance (C). Therefore, we can substitute the values to find the potential difference (V): \(V = \frac{8.0 \times 10^{-2} \mathrm{C}}{4.0 \times 10^{-5} \mathrm{F}} = 2000\ \mathrm{V}\) So the potential difference across the plates of the capacitor is \(2000\ \mathrm{V}\).