My Notas Ask Your Teach electric fleld with a magnitude of 3.40 kN/C is applied

Question

My Notas Ask Your Teach electric fleld with a magnitude of 3.40 kN/C is applied along the x axis (a) Calculate the electric flux through a rectangular plane 0.300 m wide and 0.720 m long if the plane is parallel to the yz plane N-m2yc N.m2/c N-m2c (b) Calculate the electric flux if the plane is paraliel to the xy plane. (c) Calculate the electric flux the plane contains the y axis, and its normal makes an angle of 42.ge with thor axis. y Notes Ask Your Teacha 1 points 585 24 P02 A vertical electric feld of magnitude 1.00 x 104 N/C exists above the Earth's surface on a day when a thunderstorm is brewing. A car with a rectangular size of traveling along a roadway sloping down at 7.0, Determine the electric flux through the bottom of the car 6.00 m by 3.00 m is My Notes Ask Your Teacher -12 points 585 24.P.11 The following charges are located inside a submarine: 4.00 uc, -9.00 , 27.0 uC, and-58.0 uC (a) Calculate the net electric flux through the submarine the number of electric fleld lines leaving the submarine greater than, equal to, or less than the number entering it? O equal to

Explanation / Answer

1A.

Electric flux is given by:

phi = B.A = B*A*cos theta

E = 3.4 kN/C = 3400 N/C

A = 0.300*0.720 m^2

theta = 0 deg, (since the plane is parallel to the yz plane, the normal line of the area lies parallel to the x axis)

So,

phi = 3400*0.300*0.720*cos 0 deg

phi = 734.4 N-m^2/C

1B.

In this case theta = 90 deg, (Since the normal line of the area is at a right angle to the x-axis)

Since cos 90 deg = 0, So

phi = E*A*cos 90 deg = 0 N-m^2/C

1C.

given that theta = 42 deg

So,

phi = 3400*0.300*0.720*cos 42 deg

phi = 545.76 N-m^2/C

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