PHY2323 Assignment 3 Solutions
Hello, dear friend, you can consult us at any time if you have any questions, add WeChat: daixieit
PHY2323 Assignment 3 Solutions
A:
With the provided information there are multiple equivalent pathways to determine the energy .
= ∰ ∙ = 2 =
No matter how you arrange your final equation, finding the electric field/Charge is a first easily done using Gauss with a spherical surface inside the capacitor
∯ ∗ ∙ ⃗ = ∯
The end result of this is
42 = 42 = = 1 ∗ 10−11
= =
Using the = 1 ∰ ∙ version, this is a spherical volume so dv = 2
= ∫ ∫ ∫ 20 2
step, which is
After simplifying and integrating the trivial phi and theta integrals
= ∫ = ( ( − ))
Subbing in the values of Q, a and b we find the energy in the capacitor to be = 5.5 ∗ 10−10
Using the W = version, we need to find the potential
Δ = − ∫ = ∫ = 110. 1
Subbing in the values of Q and V, we find again that = 5.5 ∗ 10−10
B:
Finding the capacitance of the capacitor is most easily done using C = Q/V, especially if you had already
found V in the previous question
The expression for Δ using the work done in the previous question is ( ( − ))
This gives us = 16022 = 9 ∗ 10−14
The other way you can calculate the capacitance of a capacitor is with the equation =
For a spherical capacitor this can be estimated to =
Unfortunately, in this case the varies by r making this equivalent to many capacitors in series, making this calculation much trickier and not recommended.
A:
This question requires the use of the Laplace equation ∇ = 0 to find the electric potential. As this is a sphere, we use the spherical laplace, and only the vector changes, giving
(2 ) = 0
To isolate V we need to integrate twice
After the first
(2 ) = 1
After the second
= − + 2
Which is the solution given in the question
We can now begin to use out boundary conditions
At r = a, V = At r = b V = 0
Using V = 0, we can find that 2 =
Using , = 0 we then have 0 = + so 1 =
So finally = 0 (1 − 1) which can be rearranged to give =
B:
= −∇
0 (1 − )
This is a spherical geometry, and only matters, so the derivative in respect to r is
=
C:
With the electric field, we can use Gauss to find the contained charge, and then the surface charge density
∯ ∗ ∙ ⃗ = ∯
The integrations just give the area of a sphere of radius a, so
42 = 2
This gives the boundary condition of an electric field near a conductor
=
=
The total charge is the charge density with the area of the inner sphere
04
( − )
D:
We have the expression for Q, the ∆ = The capacitance is simply
0 and the capacitance can be found with C = Q/V
4
( − )
When the distance between b and a is small, you can estimate this as = which is the spherical
equivalent of =
2022-04-01