A 5.80-\mu F, parallel-plate, air capacitor has a plate separation

Susan Nall

Susan Nall

Answered question

2022-01-05

A 5.80μF, 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 Jm3.

Answer & Explanation

Janet Young

Janet Young

Beginner2022-01-06Added 32 answers

Required
We are asked to find the energy density u in Jm3
Solution:
The electric potential energy stored in a charged capacitor is just equal to the amount to work required to charge it that is, to separate opposite charges and place them on different conductors. We can calculate the energy density u by calculating the electric field between the two conductors as in equation 24.11 u=12ϵE2(1)
Where the electric field E is related to the separated distance between the two conductors as next E=Vd
If we plug the expression of E into equation (1) we would get the next relation between the energy density, potential difference and the separated distance as next u=12ϵ(Vd)2(2)
Now we can plug our values for V, d and ϵ into equation (2) to get u
u=12ϵ(Vd)2
=12(8.854×1012C2Nm2)(400.0V0.005m)2
=0.028Jm3
As shown by equation (1) we can think of the energy as being stored in the field in the region between the plates.
Navreaiw

Navreaiw

Beginner2022-01-07Added 34 answers

The expression of electrostatic energy density is uE=12ϵ0E2
Here, E=Vd=4005×103=8×104Vm
Thus, uE=0.5×(8.85×1012)×(8×104)2=2.83×102Jm3
karton

karton

Expert2022-01-10Added 613 answers

The expression of electrostatic energy density is uE=12ϵ0E2
Here, E=V/d=4005×103=8×104V/m
Thus, uE=0.5×(8.85×1012)×(8×104)2=2.83×102J/m3

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