1.The electric field is measured all over the surface of a cylinder whose diamet
ID: 1493902 • Letter: 1
Question
1.The electric field is measured all over the surface of a cylinder whose diameter is 8 cm and whose height is 20 cm, as shown in the diagram. At every location on the surface the electric field points in the same direction (+y). E1 is found to be 551V/m; E2 is 765V/m; E3 is 1256V/m.
(b) What is the net electric flux on this surface?
(c) How much charge is inside the surface? 0 = 8.85e-12 C2/N m2.
2.(a) Suppose that the small area A in the figure measures 4 mm by 3 mm, and it is oriented in such a way that its outward-going normal is at an angle of 32 degrees to the direction of the electric field. What is A?(b) If the electric field is 800 volts/meter, what is the flux EA?
Explanation / Answer
A1 = pi*r^2
r = 4 cm = 0.04 m
flux through top = phi1 = E1*A1*cos0 = 551*pi*0.04^2 = 2.77 Nm^2/C
flux through side surface = phi2 = E2*A2*cos90 = 0
flux through bottom = phi3 = E3*A3*cos180 = -1256*pi*0.04^2 = -6.313 Nm^2/C
total flux = 2.77 - 6.313 = -3.543 Nm^2/C <<<------answer
+++++++++++++++++++++++++
(c)
from gauss law total flux = Qin/e
3.543 = Qin/(8.84*10^-12)
Qin = 3.13*10^-11 C
+++++++++++++++++++++++
2 (a)
A = 4*10^-3*3*10^-3*cos32 = 1.1*10^-5 m^2
(b)
flux = 800*1.1*10^-5
flux = 0.0088 Nm^2/C
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