Calculating Electric Flux and Field: Flat Sheet and Solid Sphere Examples

[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?
#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?

 

[Doc Al]

#! 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.)

 

[quicknote]

Firstly, in the calculation for electric flux through the flat sheet, the angle should be converted to radians before using it in the cosine function. So, it should be cos(20 deg) = cos(20*pi/180) = 0.9397. Also, the units for electric flux are N*m^2/C, so the final answer should be multiplied by the unit conversion factor of 1/C to get the correct unit.

For the solid sphere, the calculation for electric field is correct. However, the electric flux should be calculated as phi = E*A*cos(theta), where theta is the angle between the electric field and the normal vector to the surface of the sphere. In this case, theta = 90 degrees, so cos(theta) = 0. Therefore, the electric flux is zero at this point. This makes sense because the electric field of a point charge at a point outside its surface is zero.