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Answer: a) 73.41 10^-12 F; b)4.83* 10^3 N/C; c) 3.66 *10^3 N/C
Explanation: To solve this problem we have to consider the following: The Capacity= Charge/Potential Difference
As we know the capacity is value that depend on the geometry of the capacitor, in our case two concentric spheres.
So Potential Difference between the spheres is given by:
ΔV=-
Where E = k*Q/ r^2
so we have
then
Vb-Va=k*Q(1/b-1/a)=kQ (ab/b-a)
Finally using C=Q/ΔV=ab/(k(b-a))
To caclulate the electric firld we first obtain the charge
Q=ΔV*C=120 V*73.41 10^-12 F=8.8 10^-9 C
so E=KQ/r^2 for both values of r
r=12.8 cm ( in meters)
r2=14.7 cm
E(r1)=4.83* 10^3 N/C
E(r2)=3.66 *10^3 N/C
Answer:
ΔS=2*m*Cp*ln((T1+T2)/(2*(T1*T2)^1/2))
Explanation:
The concepts and formulas that I will use to solve this exercise are the integration and the change in the entropy of the universe. To calculate the final temperature of the water the expression for the equilibrium temperature will be used. Similarly, to find the change in entropy from cold to hot water, the equation of the change of entropy will be used. In the attached image is detailed the step by step of the resolution.
