A 5.80-uF, parallel-plate, air capacitor has a plate separation o...

Question

A 5.80-uF, parallel-plate, air capacitor has a plate separation of 5.00 mm and is charged to a potential difference of 400 V. Calculate the energy density in the region between the plates, in units of J/m^3.

Accepted Answer

Welcome back everybody. We are making observations about a battery connected to a capacitor here. Now we are told that the potential difference is 12 volts and that the plates of the capacitor are separated by one millimeter or 10.1 m. And we are tasked with finding what the energy density is stored in between the plates. Luckily we have a formula for this. This is just one half, that's the constant epsilon, not times the magnitude of the electric field squared. Now we need to know what the magnitude of the electric field is. And luckily this is just the potential difference divided by our distance. So this is just gonna be 12 divided by 120.1 giving us 1.2 times 10 to the fourth volts per meter. So now we are ready to calculate what the energy density is and this is going to be one half times or epsilon not which is equivalent to 8.854 times 10 to the negative 12 times 1.2 times 10 to the fourth squared. Which when you plug into your calculator, we get a final answer of 6.37 times 10 to the negative fourth jewels per meter cubed corresponding to our final answer choice of C. Thank you all so much for watching. Hope this video helped. We will see you all in the next one

A 5.80-uF, parallel-plate, air capacitor has a plate separation of 5.00 mm and is charged to a potential difference of 400 V. Calculate the energy density in the region between the plates, in units of J/m^3.

Watch next