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Need help with homework

by gmuniz
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gmuniz
#1
Sep18-05, 04:02 PM
P: 6
#! A flat sheet is in the shape of a rectangle with sides of length 0.400m and0.600m. the sheet is immersed in a uniform electric field 71.0 N/C that is directed at 20 deg. from the plane of the sheet. Find the magnitude of the electric flux through the sheet?
A = L*W= 0.600m*.400.
E = 71 N/C
angle = 20 deg
phi = E*A*cos(20 deg)
I don't know where i am making my mistake?
#2 a solid sphere with radius 0.460.m carries a net charge of 0.300nC. Find the magnitude of the electric field at a point 0.103 m outside the surface of the sphere?
A =4*pi*r^2= 4*pi*(0.460)^2
E =(1/4*pi*epsilon_0)*(q/r^2)= (1/(4*pi*epsilon_0))((.300*10^-9)/(0.103^2)) PHI=E*A where is my mistake?
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Doc Al
#2
Sep18-05, 04:44 PM
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Quote Quote by gmuniz
#! A flat sheet is in the shape of a rectangle with sides of length 0.400m and0.600m. the sheet is immersed in a uniform electric field 71.0 N/C that is directed at 20 deg. from the plane of the sheet. Find the magnitude of the electric flux through the sheet?
A = L*W= 0.600m*.400.
E = 71 N/C
angle = 20 deg
phi = E*A*cos(20 deg)
I don't know where i am making my mistake?
You need the angle that the field makes with the normal to the plane, which is 90-20 = 70 degrees.
#2 a solid sphere with radius 0.460.m carries a net charge of 0.300nC. Find the magnitude of the electric field at a point 0.103 m outside the surface of the sphere?
A =4*pi*r^2= 4*pi*(0.460)^2
E =(1/4*pi*epsilon_0)*(q/r^2) = (1/(4*pi*epsilon_0))((.300*10^-9)/(0.103^2)) PHI=E*A where is my mistake?
That formula is for the field from a point source (or a spherically symmetric charge distribution); the distance is measured from the center of the sphere, not the surface. That's all you need. (Assume the charge is distributed uniformly.)


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