contestada

(a) What is the energy stored in the 10.0 μF capacitor of a heart defibrillator charged to 9.00×103V ? (b) Find the amount of stored charge.

Respuesta :

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Answer:

(a) [tex]U=405J[/tex]

(b) [tex]Q=0.09C[/tex]

Explanation:

(a) The potential energy stored in a capacitor is given by the expression:

[tex]U=\frac{QV}{2}[/tex]

Q is the stored charge and V the potential difference between capacitor plates. In a capacitor, we have:

[tex]Q=CV[/tex]

Replacing this in the energy equation:

[tex]U=\frac{CV^2}{2}\\U=\frac{10*10^{-6}F(9*10^3V)^2}{2}\\U=405J[/tex]

(b) Using the energy expression and solving for Q:

[tex]U=\frac{QV}{2}\\Q=\frac{2U}{V}\\Q=\frac{2(405J)}{9*10^3V}\\\\Q=0.09C[/tex]

Cricetus

(a) The energy stored is "405 J".

(b) Amount of stored charge is "0.09 C".

(a)

As we know, the stored P.E in capacitor given by:

→ [tex]U= \frac{QV}{2}[/tex]

or,

→ [tex]Q=CV[/tex]

then,

→ [tex]U= \frac{CV^2}{2}[/tex]

By substituting the values, we get

      [tex]= \frac{10\times 10^{-6} F(9\times 10^3)^2}{2}[/tex]

      [tex]= 405 \ J[/tex]

(b)

By using the energy expression, we get

→ [tex]U = \frac{QV}{2}[/tex]

or,

→ [tex]Q=\frac{2U}{V}[/tex]

      [tex]= \frac{2(405)}{9\times 10^3}[/tex]

      [tex]= 0.09 \ C[/tex]

Thus the above answers are correct.          

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