A 5.80−μF, 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.

Question

Janet Young · Accepted Answer

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=12ϵ∘E2(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(8.854×10−12C2N⋅m2)(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 · Accepted Answer

The expression of electrostatic energy density is uE=12ϵ0E2  Here, E=Vd=4005×10−3=8×104Vm  Thus, uE=0.5×(8.85×10−12)×(8×104)2=2.83×10−2Jm3

karton · Accepted Answer

The expression of electrostatic energy density is uE=12ϵ0E2 Here, E=V/d=4005×10−3=8×104V/m Thus, uE=0.5×(8.85×10−12)×(8×104)2=2.83×10−2J/m3

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

Answered question

Answer & Explanation

New Questions in Other