2. [15 points] (a) A charge q is placed at the center of two concentric cubes as
ID: 3280293 • Letter: 2
Question
2. [15 points] (a) A charge q is placed at the center of two concentric cubes as shown. T side lengths of the smaller cube and the bigger cube are 0.500 m and 1.00 m respectively. f the electric flux .ir l tri-' /(4) what is the value ofq? one facg of the smaller cube is 1.40 x 10 N.mc, (in) what is the total flux through the bigger (b) [10 points] A charge q = 4.00 pC is placed a distance of 0.500 m above the center of a square sheet of side length 1.00 m as shown in the figure. What is the electric flux through the sheet? [No integration needed!! E-JA sh 0.500 m 100 m "()(1.Explanation / Answer
2) i) for the total flux
total flux = 6 * flux through each face
total flux = 6 * 1.40 *10^6 N.m^2/C
Using Gauss law
q/epsilon = total flux
q/(8.854 *10^-12) = 6 * 1.40 *10^6
q = 7.43 *10^-5 C
the charge is 7.43 *10^-5 C
ii) as the charge inside the larger cube is same as the smaller cube , hence , the total flux will be same.
total flux through the bigger cube = 6 * 1.40 *10^6 N.m^2/C
total flux through the bigger cube = 8.4 *10^6 N.m^2/C
b)
as we can assume that this is one face of a imaginary cube
electric flux through the sheet = q/(epsilon * 6)
electric flux through the sheet = 4 *10^-6/(6 * 8.854 * 10^-12)
electric flux through the sheet = 7.53 *10^4 N.m^2/C
the electric flux through the sheet is 7.53 *10^4 N.m^2/C
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