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never [62]
3 years ago
9

The flash unit in a camera uses a 3.0 V battery to charge acapacitor. The capacitor is then discharged through a flashlamp.The d

ischarge takes 7 µs, andthe average power dissipated in the flashlamp is 8 W. What is the capacitance of the capacitor?
Physics
1 answer:
Varvara68 [4.7K]3 years ago
6 0

Answer:

C=2.54\cdot 10^5\ F

Explanation:

<u>Energy Stored on a Capacitor</u>

This energy is stored in the electric field present when a voltage V is applied to a capacitor C. It can be computed by the formula

\displaystyle  E=\frac{CV^2}{2}

We can compute the energy by knowing the power dissipated in the flashlamp is P=8 W in a time of 7 \mu s=7\cdot 10^{-6}\ sec

\displaystyle  E=\frac{P}{t}=\frac{8\ W}{7\cdot 10^{-6}\ sec}=1.14\cdot 10^6\ J

Solving the first equation for C

\displaystyle  C=\frac{2E}{V^2}

\displaystyle  C=\frac{2\times 1.14\cdot 10^6}{3^2}

\boxed{C=2.54\cdot 10^5\ F}

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The ground state energy of an oscillating electron is 1.23 eV. How much energy must be added to the electron to move it to the t
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Answer:

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  • The energy that must be added to the electron to move it to the fourth excited state is  -1.181 eV

Explanation:

Given;

Energy of electron in ground state (n = 1 ) = 1.23 eV

E₁ = 1.23 eV

Eₙ = E₁ /n²

where;

E₁ is the energy of the electron in ground state

n is the energy level,

For third excited state, n = 4

E₄ = E₁ /4²

E₄ = (1.23 eV) / 16

E₄ = 0.077 eV

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For fourth excited state, n = 5

E₅ = E₁ /5²

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The energy that must be added to the electron to move it to the fourth excited state is  -1.181 eV

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