A rectangular box has the following dimensions [Length (x-axis)= 10 cm, width (z

Question

A rectangular box has the following dimensions [Length (x-axis)= 10 cm, width (z- axis= 15 cm and height (y-axis)= 5 cm. It is placed with its front bottom left corner at the origin as shown in the figure. An uniform Electric Field having a magnitude E=3i-6j+5k (N/c) is acting on the box. a) Find the total flux on S-Front, ------(5) (b) Find the total flux on S-Top,------------- (5) (c) Find total flux on S-right side,----------------- (5) (d) The total flux for the entire cube (i.e. sum of all 6 surfaces) ------(10)

Explanation / Answer

Flux = Electric field*Area

Now E = 3i-6j+5k

1) Flux on S-front

=  Electric field*Area = (3i-6j+5k)*(Area in x-direction)=(3i-6j+5k)*(Ai) = 3A = 3*(y*z) = 3*(0.05*0.15) = 0.0225 Nm^2/C

2)

Flux on S-top

=  Electric field*Area = (3i-6j+5k)*(Area in z-direction)=(3i-6j+5k)*(Ak) = 5A = 5*(x*y) = 5*(0.10*0.05) = 0.0250 Nm^2/C

3)

Flux on S-right side

=  Electric field*Area = (3i-6j+5k)*(Area in y-direction)=(3i-6j+5k)*(Aj) = - 6A = -6*(x*y) = -6*(0.10*0.15) = - 0.09 Nm^2/C

4) Net electric flux through all six side is zero, because from each pair of sides equal and opposite electric flux is there. e.g. flux through S-top and S-down is equal and opposite so cancelled.

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