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[SOLVED] Flux through a surface
The square surface shown in Figure 24-26 measures 3.8 mm on each side. It is immersed in a uniform electric field with magnitude |E| = 2000 N/C. The field lines make an angle of 35° with a normal to the surface, as shown. Take that normal to be directed "outward," as though the surface were one face of a box. Calculate the electric flux through the surface.
http://personalpages.tds.net/~locowise/test/24-26.gif
E=2000N/C
theta=(180-35)=145
A=(0.0038m)^2=1.444*10^-5 m^2
flux=EAcos(theta)
flux=2000N/C * (1.444*10^-5 m^2) cos145 = -0.023657 N*m^2/C
Homework Statement
The square surface shown in Figure 24-26 measures 3.8 mm on each side. It is immersed in a uniform electric field with magnitude |E| = 2000 N/C. The field lines make an angle of 35° with a normal to the surface, as shown. Take that normal to be directed "outward," as though the surface were one face of a box. Calculate the electric flux through the surface.
http://personalpages.tds.net/~locowise/test/24-26.gif
E=2000N/C
theta=(180-35)=145
A=(0.0038m)^2=1.444*10^-5 m^2
Homework Equations
flux=EAcos(theta)
The Attempt at a Solution
flux=2000N/C * (1.444*10^-5 m^2) cos145 = -0.023657 N*m^2/C
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