The figure shows a box on whose surfaces the electric field is measured to be ho

Question

The figure shows a box on whose surfaces the electric field is measured to be horizontal and to the right. On the left face (3 cm by 2 cm) the magnitude of the electric field E1 is 400 volts/meter, and on the right face the magnitude of the electric field E2 is 1000 volts/meter. On the other faces only the direction is known (horizontal). Calculate the electric flux on every face of the box, the total flux, and the total amount of charge that is inside the box. E2 2 cm 8 cm Flux through top Vm Flux through bottom Vm Flux through left Vm Flux through right Flux through front Flux through back Total Flux Vm Vm Vm Vm Total charge inside box (recall co-8.85 x 10-12 c2/N-m2)

Explanation / Answer

Electric flux = E*A*costheta

E = magnitude of electric field

A = area of surface

theta = angle between A and E

flux through top = Etop*A*cos90 = 0

flux through bottom = Ebottom*A*cos90 = 0

flux through left = E1*A*cos180 = 400*0.02*0.03*cos180 = -0.24 Vm

flux through right = E2*A*cos0 = 1000*0.02*0.03*cos0 = 0.6 Vm

flux through front = Efront*A*cos90 = 0

flux through back = Eback*A*cos90 = 0

Total flux = 0.6-0.24 = 0.36 Vm

from Gauss law

Total flux = charge inside/eo

eo = permitivitty of free space = 8.85*10^-12

Total charge inside = 0.36*8.85*10^-12 = 3.186*10^-12 C

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