In problem 14 what is the charge/area on the inner surface of the hollow sphere
ID: 1555021 • Letter: I
Question
In problem 14 what is the charge/area on the inner surface of the hollow sphere on c/m^2? a) 3.98 b) 10.61 c) 7.96 d) 2.55 In problem 14 the small conducting sphere is brought in contact with inner surface of the hollow sphere. What is the charge/area on the outer surface of the sphere (in C/m^2)? a) -0.221 b) 0.318 c) -0.497 d) -0.589 In problem 16, what is the charge/area on the inner surface of the hollow sphere (in c/m^2)? a) 0 b) -0.447 c) 0.589 d) 0.318 Two parallel conducting sheets of area A are separated by a gap distance h. They are connected to of voltage V. Let A = 2.00 m^2, d = 0.00300m, and V = 150 V. Find the electric field in the gap region (in V/m). a) 50,000 b) 75,000 c) 57, 100 d)65, 400 In problem 18, find the energy stored between the conducting sheets (in J) a) 1.48 times 10^-4 b) 1.01 times 10^-4 c) 2.09 times 10^-4 d) 6.64 times 10^-5 If the battery is removed and the plates are connected to a capacitor with capacitance C_2 = 7.00 times 10^-9 F. Find the new value for the energy stored between the conducting sheets (in J). a) 2.66 times 10^-5 b) 1.39 times 10^-5 c) 7.60 times 10^-5 d) 4.64 times 10^-5Explanation / Answer
a)
E=V/d=150/0.003=50000 V/m
b)
C=eoA/d=(8.8542*10-12)*2/0.003=5.9*10-9 F
energy stored
U=(1/2)CV2=(1/2)*(5.9*10-9)*1502
U=6.64*10-5 J
c)
initial charge stored
Q=CV=(5.9*10-9)*150=8.85*10-7 J
equivalent capacitance
Ceq=(5.9+7)*10-9=1.29*10-8 F
new potential difference across capacitors
Vnew=Q/Ceq=(8.85*10-12)/(1.29*10-8)=68.6 Volts
energy stored in plates
Unew=(1/2)(5.9*10-9)*68.62=1.39*10-5 J