56.
Electric flux through a enclosed surface is given by:
Electric flux = Q_enclosed/e0
Q_enclosed = net charge inside the surface = 2.0*10^-6 C
e0 = 8.85*10^-12
Now from above equation we can see that electric flux does not depend on the size of enclosed surface, and since net charge inside all three surface is same, so electric flux through all three surface will also be same. which will be
Electric flux = (2*10^-6)/(8.85*10^-12)
Electric flux = 2.25988*10^5 N-m^2/C = 2.3*10^5 N-m^2/C
Part A
Electric flux = 2.3*10^5 N-m^2/C
Part B
Electric flux = 2.3*10^5 N-m^2/C
Part C
Electric flux = 2.3*10^5 N-m^2/C
57.
Electric flux in external electric field is given by:
Electric flux () = E.A = E*A*cos
cos = /(E*A)
Given values are
Electric flux = = 78 N-m^2/C
Electric field = E = 1.44*10^4 N/C
A = Area of circular surface = pi*r^2
r = radius = 0.057 m
A = pi*(0.057)^2
So,
cos = /(E*A)
= arccos [78/(1.44*10^4*pi*0.057^2)]
= 57.9 deg = Angle between direction of E and normal to the surface
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