Which tools would be necessary to determine whether or not a large block will float, without using water?
1 answer:
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Part A)
First of all, let's convert the radii of the inner and the outer sphere:
The capacitance of a spherical capacitor which consist of two shells with radius rA and rB is
Then, from the usual relationship between capacitance and voltage, we can find the charge Q on each sphere of the capacitor:
Now, we can find the electric field at any point r located between the two spheres, by using Gauss theorem:
from which
In part A of the problem, we want to find the electric field at r=11.1 cm=0.111 m. Substituting this number into the previous formula, we get
And so, the energy density at r=0.111 m is
Part B) The solution of this part is the same as part A), since we already know the charge of the capacitor:
. We just need to calculate the electric field E at a different value of r: r=16.4 cm=0.164 m, so
And therefore, the energy density at this distance from the center is
First of all, let's convert the radii of the inner and the outer sphere:
The capacitance of a spherical capacitor which consist of two shells with radius rA and rB is
Then, from the usual relationship between capacitance and voltage, we can find the charge Q on each sphere of the capacitor:
Now, we can find the electric field at any point r located between the two spheres, by using Gauss theorem:
from which
In part A of the problem, we want to find the electric field at r=11.1 cm=0.111 m. Substituting this number into the previous formula, we get
And so, the energy density at r=0.111 m is
Part B) The solution of this part is the same as part A), since we already know the charge of the capacitor:
And therefore, the energy density at this distance from the center is
Answer:
Part(a): the capacitance is 0.013 nF.
Part(b): the radius of the inner sphere is 3.1 cm.
Part(c): the electric field just outside the surface of inner sphere is .
Explanation:
We know that if 'a' and 'b' are the inner and outer radii of the shell respectively, 'Q' is the total charge contains by the capacitor subjected to a potential difference of 'V' and '' be the permittivity of free space, then the capacitance (C) of the spherical shell can be written as
Part(a):
Given, charge contained by the capacitor Q = 3.00 nC and potential to which it is subjected to is V = 230V.
So the capacitance (C) of the shell is
Part(b):
Given the inner radius of the outer shell b = 4.3 cm = 0.043 m. Therefore, from equation (1), rearranging the terms,
Part(c):
If we apply Gauss' law of electrostatics, then