A 4.03 x 10-5 F capacitor stores 5.15 x 10-3 J of energy. What is the voltage across the capacitor?​

V = 0.06255 Volts

Explanation:

Given the following data;

Capacitance = 4.03 * 10^{-5} F

Energy = 5.15 * 10^{-3} J

To find the voltage across the capacitor;

Mathematically, the energy stored in a capacitor is given by the formula;

E = ½cv²

Where;

C is the capacitor

E is the energy stored

v is the voltage.

Substituting into the formula, we have;

5.15 * 10^{-3} = ½ * 4.03 * 10^{-5} * v²

0.00515 = 0.00002015 * v²

V² = 0.00002015/0.00515

V² = 0.003912

Taking the square root of both sides, we have;

V = √0.003912

V = 0.06255 Volts

beta is electron from nucleus

something from nucleus has changed, but charge must be same for cons of charge

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average time interval starts degrading is the segment where the car is slowing down