Two charges of 15 pc and -40 pC are inside a cube with sides that are of of 0.40

Question

Two charges of 15 pc and -40 pC are inside a cube with sides that are of of 0.40.m length. Determine the net electric flux through the surface of the cube. +2.8 N m^2/C +1.1 N. m^2/c -2.8 N m^2/C -0.47 N m^2/C The total electric flux through a closed cylindrical (length = 1.2 m, diameter = 0.20 m) surface is equal to -5.0 N.m^2/C. Determine the net charge within the cylinder. -62 pC 53 pC -44 pC 71 pC 16 pC A uniform linear charge density of 40 nC/m is distributed along the entire x axis. Consider a spherical (radius = 5.0 cm) surface centered on the origin. Determine the electric flux through this surface. - 68 N m^2/C 62 N m^2/C 45 N m^2/C 79 N. m^2/C 23 N.m^2/c A long nonconducting cylinder (radius = 12 cm) has a charge of uniform density (5.0 nC/m^3) distributed a throughout its volume. Determine the magnitude of the electric field 15 cm from the axis of the cylinder 20 N/C 27 N/C 16 N/C 12 N/C 54 N/C A 3.5-cm radius hemisphere contains a total charge of 6.6 times 10^-7 C. The flux through the rounded portion of the surface is 9.8 times 10^4 N.m^4/C. The flux through the flat base is: 0 N.m^2/C +2.3 times 10^4 N.m^2/C -9.8 times 10^4 N. m^2/C +9.8 times 10^4 N m^2/C If the electric field just outside a thin conducting sheet is equal to 1.5 N/C, determine the surface charge density on the conductor. 53 pC/m^2 27 pC/m^2 35 pC/m^2 13 pC/m^2 6.6 pC/m^2

Explanation / Answer

here,

q1 = 15 pC

q2 = - 40 pC

the net charge enclosed in the cube , q = q1 + q2 = - 25 pC

q = - 25 * 10^-12 C

using Gauss's law

the electric flux through the cube , phi = q/e0

phi = - 25 * 10^-12 /(8.85 * 10^-12)

phi = - 2.8 N.m^2 /C

so the electric flux through the surface of cube is d) - 2.8 N.m^2 /C

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