The electronic flash attachment for a camera contains a capacitor for storing the energy used to produce the flash. In one such unit, the potential difference between the plates of a 775 µF capacitor is 330 V.

(a) Determine the energy that is used to produce the flash in this unit.

(b) Assuming that the flash lasts for 5.0*10^-3 s, find the effective power or "wattage" of the flash.

So far:
0.000775 Farads= Coulumbs / 330V
Coulumbs=0.25575C

Now here I'm not sure what to do next.

But I push on:
1A=1C/s

So:
0.25575C/5x10^-3s=51.15A

P=IV

P=51.15A*330V=16879.5W

But, that answer is kicked backed as wrong, and I'm not quite sure how to go about part A.

(a) Energy stored = (1/2)CV^2

(b) Power = Energy/(Flash duration)

Well, let's take a humorous approach to solve this problem!

(a) To determine the energy used to produce the flash, we need to use a capacitor. Ah, the capacitor – it's like the very enthusiastic friend who stores all the excitement and energy for us!

Now, the potential difference between the plates of our 775 µF capacitor is 330 V. Well, that's quite a shocking potential difference, isn't it?

To calculate the energy, we can use the formula:

Energy (E) = (1/2) * C * V^2

Let's plug in the values and calculate!

E = (1/2) * 775 µF * (330 V)^2

But wait, we have some funny units here! Who measures energy in microFarads and volts? So, let's convert them into more friendly units!

775 µF = 775 * 10^(-6) F
330 V = 330 J/C (because V = J/C)

E = (1/2) * 775 * 10^(-6) F * (330 J/C)^2

Now, let's calculate that energy! (Feel free to use a calculator)

(b) We are now assuming that the flash lasts for 5.0*10^(-3) s. So, it's like a quick burst of light that's gone in the blink of an eye!

To find the effective power or "wattage" of the flash, we can use the formula:

Power (P) = Energy (E) / Time (t)

But wait, we already calculated the energy in part (a), so we just need to divide it by the given time of 5.0*10^(-3) s.

P = E / t

Again, feel free to use a calculator to find the answer.

And voila! You have the energy used to produce the flash and the power or "wattage" of the flash. Keep shining bright like a camera flash!

To determine the energy used to produce the flash in this unit, you can use the formula:

Energy (Joules) = (1/2) * C * V^2

(a) Plug in the values for capacitance (C) and potential difference (V) into the formula:

Energy = (1/2) * (775 µF) * (330 V)^2

Convert the capacitance from microfarads (µF) to farads (F):
775 µF = 775 x 10^-6 F

Energy = (1/2) * (775 x 10^-6 F) * (330 V)^2

Now, calculate the energy.

(b) To find the effective power or "wattage" of the flash, you can use the formula:

Power (Watts) = Energy / Time

However, before using this formula, make sure that the energy value is in joules and the time value is in seconds.

Since the energy was already calculated in joules in part (a), you can directly use it.

Given that the flash lasts for 5.0 * 10^-3 s, plug in the values into the formula:

Power = Energy / Time
Power = Energy / (5.0 * 10^-3 s)

Now, calculate the power.

To determine the energy used to produce the flash in the camera's flash unit, you can use the formula for the energy stored in a capacitor:

E = (1/2) * C * V^2

where E is the energy in joules, C is the capacitance in farads, and V is the potential difference (voltage) in volts.

In this case, you already have the capacitance (C) and the potential difference (V).

(a) Plugging in the given values, the energy can be calculated as:

E = (1/2) * (775 * 10^(-6)) * (330)^2

E ≈ 42.75725 joules

So, the energy used to produce the flash in this unit is approximately 42.76 joules.

Moving on to part (b), to find the effective power or "wattage" of the flash, you can use the formula:

P = E / t

where P is the power in watts, E is the energy in joules, and t is the time in seconds.

Given that the flash lasts for 5.0 * 10^(-3) seconds, you can calculate:

P = 42.75725 joules / (5.0 * 10^(-3)) s

P ≈ 8551.45 watts

Therefore, the effective power or "wattage" of the flash is approximately 8551.45 watts.