here,
Q1 = - 1 uC = - 1 * 10^-6 C
Q2 = 3 uC = 3 * 10^-6 C
Q3 = - 2 uC = - 2 * 10^-6 C
Q4 = 1 uC = 1 * 10^-6 C
using Gauss's law
the electric flux through the surfacr below , phi = charge enclosed /e0
phi = (Q1 + Q2)/e0
phi = (-1 + 3) * 10^-6 /(8.85 * 10^-12)
phi = 2.26 * 10^5 N.m^2 /C
24. What is the electric flux through the surface below? Q1--1pC, Q2 3pC , Q3--ZuC. 03...
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Q3 q1 q2 [In all questions, if a variable isn't needed, you may omit it. 1. What is the electric potential, at the point where qi is located, due to Qs and q2? 2. What is the electric field, at the point where qi is located, due to Qs and q2? Provide magnitude and direction. 3. In terms of qi, r, and R, what is the direction and magnitude of the electrostatic force on qı? Assume q2-2qi, but Q3 -3qi...
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Course Contents » ... » Homework 03 » Flux Above Surface An electric flux of 159 N.m2/C passes through a flat horizontal surface that has an area of 0.77 m2. The flux is due to a uniform electric field. What is the magnitude of the electric field if the field points 15o above the horizontal?
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