When 16.6 V is applied to a capacitor, it stores an energy of [tex]$8.15 \times 10^{-4} J$[/tex]. How much charge is on the plates?

[tex][?] \times 10^{[?]} C[/tex]



Answer :

Sure, let's solve this step-by-step!

Given:
- Voltage ([tex]\(V\)[/tex]) = 16.6 volts
- Energy ([tex]\(E\)[/tex]) = [tex]\(8.15 \times 10^{-4}\)[/tex] joules

We need to find:
- Charge ([tex]\(Q\)[/tex]) on the plates in the form [tex]\(n \times 10^{m}\)[/tex] coulombs (C)

### Step-by-Step Solution

1. Identify the Formula to Calculate Capacitance ([tex]\(C\)[/tex]):
The energy stored in a capacitor is given by the formula:
[tex]\[ E = \frac{1}{2} C V^2 \][/tex]
where:
- [tex]\(E\)[/tex] is the energy in joules,
- [tex]\(C\)[/tex] is the capacitance in farads, and
- [tex]\(V\)[/tex] is the voltage in volts.

2. Rearrange the Formula to Solve for Capacitance ([tex]\(C\)[/tex]):
[tex]\[ C = \frac{2E}{V^2} \][/tex]

3. Substitute Given Values into the Capacitance Formula:
[tex]\[ C = \frac{2 \times 8.15 \times 10^{-4}}{(16.6)^2} \][/tex]

4. Result for Capacitance:
After substituting the values and performing the calculation, we get:
[tex]\[ C \approx 5.91522717375526 \times 10^{-6} \, \text{farads (F)} \][/tex]

5. Identify the Formula to Calculate Charge ([tex]\(Q\)[/tex]):
[tex]\[ Q = C \times V \][/tex]
where:
- [tex]\(Q\)[/tex] is the charge in coulombs (C),
- [tex]\(C\)[/tex] is the capacitance in farads (F),
- [tex]\(V\)[/tex] is the voltage in volts (V).

6. Substitute the Capacitance and Voltage into the Charge Formula:
[tex]\[ Q = 5.91522717375526 \times 10^{-6} \times 16.6 \][/tex]

7. Result for Charge:
After performing the calculation, we get:
[tex]\[ Q \approx 9.819277108433733 \times 10^{-5} \, \text{coulombs (C)} \][/tex]

Thus, the charge on the plates is approximately:
[tex]\[ Q \approx 9.819277108433733 \times 10^{-5} \, \text{C} \][/tex]

This can be expressed as:
[tex]\[ Q \approx 9.82 \times 10^{-5} \, \text{C} \][/tex]

So, when 16.6 V is applied to the capacitor, and it stores an energy of [tex]\(8.15 \times 10^{-4}\)[/tex] joules, the charge on the plates is approximately:
[tex]\[ 9.82 \times 10^{-5} \, \text{C} \][/tex]