Question

9) A non-uniform electric field is directed along the x-axis at all points in space. This magnitude of the field varies with x, but not with respect to y or z. The axis of a cylindrical surface, 0.80 m long and 0.26 m in diameter, is aligned parallel to the x-axis, as shown in the figure. The electric fields E1 and E2, at the ends of the cylindrical surface, have magnitudes of 6930 N/C and 710 N/C respectively, and are directed as shown. What is the net electric flux passing through the cylindrical surface? What charge must be inside the cylinder? E. -0.80 m

Answer 1

vert A vector_1 vert = vert A vector_2 = pi r^2 = pi d^2/4 = pi (0.26)^2/4 = 0.053066 m^2 Since, A vector_1 and E vector_1 are opposite in direction. Hence, phi_1 = E vector_1 middot A vector_1 = E_1 A_1 cos 180 degree = -E_1 A_1 = -(6930) (0.053066) = -367.747 N - m^2/C Now, A vector_2 and E vector_2 are in same direction. Hence, phi_2 = E vector_2 middot A vector_2 = E_2 A_2 cos 180 degree = E_2 A_2 = (710) (0.053066) 37.6768 N - m^2/C Therefore Net flux, phi_net = -367.747 + 37.6768 = -330.07014 N - m^2/C
Electric flux is equal to scalar product of electric field and area vector.Net electric flux obtain from algebraic sum of flux linked to surfaces of cylinder.

Enclosed charge can be obtain from gauss law.