Question

A 4.00 -pF parallel plate capacitor has a potential difference of \(10.0 \mathrm{~V}\) across it. The plates are \(3.00 \mathrm{~mm}\) apart, and the space between them contains air. a) What is the charge on the capacitor? b) How much energy is stored in the capacitor? c) What is the area of the plates? d) What would the capacitance of this capacitor be if the space between the plates were filled with polystyrene?

Short answer

Question: Calculate the charge, stored energy, area of the plates, and the capacitance of a parallel plate capacitor when it has a capacitance of 4.00 pF, a potential difference of 10.0 V, a distance between the plates of 3.00 mm, and if the space between the plates contains polystyrene material. Answer: The charge on the capacitor is \(4.00 \times 10^{-11}\) coulombs, the energy stored in the capacitor is \(2.00 \times 10^{-10}\) joules, the area of the plates is \(1.18 \times 10^{-6}\) square meters, and the capacitance with polystyrene is \(10.4 \text{ pF}\).

Step-by-step solution

a) Charge on the capacitor

To find the charge, we use the formula \(Q = CV\). We know the capacitance, \(C\), is \(4.00\) pF and the potential difference, \(V\), is \(10.0\) V. First, we have to convert the capacitance to Farads: \(4.00\) pF = \(4.00 \times 10^{-12}\) F. Now we can calculate the charge: $$Q = (4.00 \times 10^{-12}\text{ F})(10.0\text{ V}) = 4.00 \times 10^{-11}\text{ C}$$ The charge on the capacitor is \(4.00 \times 10^{-11}\) coulombs.

b) Energy stored in the capacitor

To find the stored energy, we use the formula \(U=\frac{1}{2}CV^2\). We have the capacitance, \(C\), as \(4.00 \times 10^{-12} \text{ F}\) and the potential difference, \(V\), as \(10.0 \text{ V}\). $$U = \frac{1}{2}(4.00 \times 10^{-12}\text{ F})(10.0\text{ V})^2 = 2.00 \times 10^{-10}\text{ J}$$ The energy stored in the capacitor is \(2.00 \times 10^{-10}\) joules.

c) Area of the plates

To find the area of the plates, we use the formula \(A = \frac{C\epsilon_0}{V}d\). We have the capacitance, \(C\), as \(4.00 \times 10^{-12} \text{ F}\), the distance between the plates, \(d\), is \(3.00 \mathrm{~mm} = 3.00 \times 10^{-3}\mathrm{~m}\), and the potential difference, \(V\), is \(10.0 \text{ V}\). The vacuum permittivity constant, \(\epsilon_0\), is \(8.85 \times 10^{-12} \text{F}\cdot\text{m}^{-1}\). $$A = \frac{(4.00 \times 10^{-12}\text{ F})(8.85 \times 10^{-12}\text{ F}\cdot\text{m}^{-1})}{(10.0\text{ V})(3.00 \times 10^{-3}\mathrm{~m})} = 1.18 \times 10^{-6} \text{m}^2$$ The area of the plates is \(1.18 \times 10^{-6}\) square meters.

d) Capacitance with polystyrene

To find the new capacitance when the space contains polystyrene, we use the formula \(C_2 = KC_1\). We have the initial capacitance, \(C_1\), as \(4.00 \times 10^{-12} \text{ F}\), and the dielectric constant of polystyrene, \(K\), is approximately \(2.6\). $$C_2 = (2.6)(4.00 \times 10^{-12}\text{ F}) = 10.4 \times 10^{-12}\text{ F}$$ The capacitance of this capacitor with polystyrene is \(10.4 \text{ pF}\).