What is the voltage across a 4.5-mu F capacitor if the charge on one plate is 0.12 mC? How much energy is stored?

To find the voltage across the capacitor, we can use the formula:<br /><br />\[ V = \frac{Q}{C} \]<br /><br />where:<br />- \( V \) is the voltage across the capacitor,<br />- \( Q \) is the charge on one plate of the capacitor,<br />- \( C \) is the capacitance of the capacitor.<br /><br />Given:<br />- \( Q = 0.12 \, \text{mC} = 0.12 \times 10^{-3} \, \text{C} \)<br />- \( C = 4.5 \, \mu\text{F} = 4.5 \times 10^{-6} \, \text{F} \)<br /><br />Substitute the given values into the formula:<br /><br />\[ V = \frac{0.12 \times 10^{-3}}{4.5 \times 10^{-6}} \]<br /><br />\[ V = \frac{0.12}{4.5} \times 10^{3} \]<br /><br />\[ V = 0.02667 \times 10^{3} \]<br /><br />\[ V = 26.67 \, \text{V} \]<br /><br />So, the voltage across the capacitor is 26.67 V.<br /><br />Next, to find the energy stored in the capacitor, we can use the formula:<br /><br />\[ E = \frac{1}{2} C V^2 \]<br /><br />Substitute the given values into the formula:<br /><br />\[ E = \frac{1}{2} \times 4.5 \times 10^{-6} \times (26.67)^2 \]<br /><br />\[ E = \frac{1}{2} \times 4.5 \times 10^{-6} \times 707.11 \]<br /><br />\[ E = 2.25 \times 10^{-6} \times 707.11 \]<br /><br />\[ E = 1.589 \times 10^{-3} \, \text{J} \]<br /><br />So, the energy stored in the capacitor is 1.589 mJ.