Question

What is the energy stored in the 10.0 μF capacitor of a heart defibrillator charged to 9.00×10^3 V?

a) 4.50 J
b) 0.45 J
c) 2.25 J
d) 20.25 J
Find the amount of stored charge.

a) 90.0 μC
b) 810 μC
c) 9.00 C
d) 1.00 μC

Answer 1

The energy stored in the 10.0 μF capacitor of a heart defibrillator charged to 9.00×10³ V is approximately 0.45 J, and the amount of stored charge is 90.0 μC.

Step-by-step explanation:

The energy (U) stored in a capacitor is given by the formula U = 1/2 * C * V², where C is the capacitance and V is the voltage across the capacitor.

In this case, the capacitor has a capacitance of 10.0 μF (which is 10.0 x 10⁻⁶ F) and is charged to a voltage of 9.00 x 10³ V.

Using the formula, we can calculate the energy stored as U = 1/2 * 10.0 x 10⁻⁶ F * (9.00 x 10³ V)² = 0.405 J, which approximately is 0.45 J (option b).

To find the amount of charge (Q) stored, we use the relationship Q = C * V.

Therefore, Q = 10.0 μF * 9.00 x 10³ V = 90.0 x 10⁻⁶ C = 90.0 μC (option a).