A capacitor has a charge of 3.450 nC and a potential difference of 5.10 V. (a) How much energy is stored in the capacitor? What is its capacitance?
The correct answer and explanation is:
To solve this, we use two key capacitor formulas:
Relevant Formulas:
- Energy stored in a capacitor:
U=12CV2U = \frac{1}{2} C V^2
- Capacitance:
C=QVC = \frac{Q}{V}
Given:
- Charge, Q=3.450 nC=3.450×10−9 CQ = 3.450 \, \text{nC} = 3.450 \times 10^{-9} \, \text{C}
- Voltage, V=5.10 VV = 5.10 \, \text{V}
(a) Capacitance:
C=QV=3.450×10−9 C5.10 V=6.765×10−10 F=676.5 pFC = \frac{Q}{V} = \frac{3.450 \times 10^{-9} \, \text{C}}{5.10 \, \text{V}} = 6.765 \times 10^{-10} \, \text{F} = 676.5 \, \text{pF}
(b) Energy Stored:
U=12CV2=12(6.765×10−10)(5.10)2U = \frac{1}{2} C V^2 = \frac{1}{2} (6.765 \times 10^{-10}) (5.10)^2 U=0.5×6.765×10−10×26.01≈8.79×10−9 J=8.79 nJU = 0.5 \times 6.765 \times 10^{-10} \times 26.01 \approx 8.79 \times 10^{-9} \, \text{J} = 8.79 \, \text{nJ}
✅ Final Answers:
- Capacitance: 676.5 pF\boxed{676.5 \, \text{pF}}
- Energy Stored: 8.79 nJ\boxed{8.79 \, \text{nJ}}
Explanation (300 words):
Capacitors are devices that store electrical energy by accumulating charge on two conductive plates separated by an insulator. The key quantities that define a capacitor’s behavior are the amount of charge it holds (QQ), the voltage across its plates (VV), and its capacitance (CC), which is a measure of how much charge it can store per volt.
To find the capacitance, we used the formula C=Q/VC = Q/V, which directly links charge and voltage. By dividing the given charge (3.450 nC) by the voltage (5.10 V), we found that the capacitance is approximately 676.5 picofarads (pF). This indicates that the capacitor can store about 676.5 picocoulombs per volt applied.
Next, to determine the energy stored, we applied the formula U=12CV2U = \frac{1}{2} C V^2. This equation shows that energy depends on both how much capacitance the device has and the square of the voltage applied. Substituting our previously found capacitance and the given voltage, we calculated the energy stored to be about 8.79 nanojoules (nJ). This amount of energy is relatively small, which is expected for a capacitor operating at nanocoulomb-level charge.
This analysis is essential in circuits where capacitors are used for energy storage, filtering, and signal timing, as it helps designers predict behavior under different voltage conditions.