Please help if you can.
1 answer:
Initially there were 10 bulbs of 60 Watt power
So total power of all bulbs = 60 * 10 = 600 W
now each bulb used for 4 hours daily
so total energy consumed daily
now we have total power consumed in 1 year
cost of electricity = 10 cents/ kWh
so total cost of energy for one year
Now if all 60 Watt bulbs are replaced by 30 Watt bulbs
So total power of all bulbs = 30 * 10 = 300 W
now each bulb used for 4 hours daily
so total energy consumed daily
now we have total power consumed in 1 year
cost of electricity = 10 cents/ kWh
so total cost of energy for one year
total money saved in 1 year
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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