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A 0.50 muF capacitor is connected to a 12 V battery. Use the expression PE = 1352_files/i0130000.jpgC(DV)1352_files/i0130001.jpg to determine how much electrical potential energy is stored in the capacitor. a. 3.0 × 101352_files/i0130002.jpg J c. 1.0 × 101352_files/i0130003.jpg J b. 6.0 × 101352_files/i0130004.jpg J d. 3.6 × 101352_files/i0130005.jpg J

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skyluke89
The capacity of the capacitor is
[tex]C=0.50 \mu F=0.50 \cdot 10^{-6} F[/tex]
and the potential difference applied across the capacitor is
[tex]V=12 V[/tex]

Therefore we can calculate the electrical energy stored in the capacitor, which is given by the following formula:
[tex]U= \frac{1}{2}CV^2 = \frac{1}{2}(0.50 \cdot 10^{-6} F)(12 V)^2=3.6 \cdot 10^{-5} J [/tex]
Lanuel

The quantity of electrical potential energy that is stored in this capacitor is equal to [tex]3.6 \times 10^{13}\;Joules[/tex]

Given the following data:

Capacitance = 0.50 muF.

Potential difference (voltage) = 12 Volts.

How to calculate the electrical potential energy.

Mathematically, the electrical potential energy stored in a capacitor is given by this formula:

[tex]E=\frac{1}{2} cv^2\\\\[/tex]

Where:

  • c is the capacitance.
  • V is the potential difference (voltage).

Substituting the given parameters into the formula, we have;

[tex]E=\frac{1}{2} \times 0.5 \times 10^{-6} \times 12^2\\\\E=3.6 \times 10^{13}\;Joules[/tex]

Read more on potential energy here: https://brainly.com/question/1242059

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