Question

Please show your work in details, write formulas clearly, and plug in values properly.

20. A cube of edge L = 10 cm has charge inside it & is located as shown in figure. The electric field given by: Ans: 26.6 nC E = (2 + 302x105 k N/C Calculate the amount ofcharge inside the cube. AZ

Answer 1

Since electric field is in z direction so flux is only from top and bottom surfaces for both surface (z = 0) e vector = 2 Times 10^5 k^n/c, a vector = (1)^2 (- k^) = -.01 k^therefore flux phi_a = (2 Times 10^5 k^) (- 0.1^2 k^) phi_b = -2 Times 1063 nm^2/c for top surface (z = .1 m) e vector = (2 + 30 Times 0.1) Times 10^5 k^= 5 Times 10^5 k^a vector = .01 k^m^2 therefore phi_t = e vector a vector = (5 Times 10^5 k^) (.01 k^) phi_t = 5 Time 1063 nm^2/c now by gauss law phi_total = phi_t + phi_b = q_in/ implies q_in = (5 - 2) Times 10^3 Times 8.85 Times 10^-12 implies q_in = 26.55 Times 10^-9 c