Question

56. A surface completely surrounds a +2.0 × 10-6 C charge. Find the electric flux through this surface when the surface is (a) a sphere with a radius of 0.50 m, (b) a sphere with a radius of 0.25 m, and (c) a cube with edges that are 0.25 m long. 57. A circular surface with a radius of 0.057 m is exposed to a uniforrm external electric field of magnitude 1.44 × 104 N/C. The magnitude of the electric flux through the surface is 78 N m2/C. What is the angle (less than 90°) between the direction of the electric field and the normal to the surface?

Answer 1

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 ($\phi$) = E.A = E*A*cos $\theta$

cos $\theta$ = $\phi$/(E*A)

Given values are

Electric flux = $\phi$ = 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 $\theta$ = $\phi$/(E*A)

$\theta$ = arccos [78/(1.44*10^4*pi*0.057^2)]

$\theta$ = 57.9 deg = Angle between direction of E and normal to the surface

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