Assignment I Begin Date: 120 2018 12:00:00 AM- Due Date: 2/3/2018 11:59:00 PM En

Question

Assignment I Begin Date: 120 2018 12:00:00 AM- Due Date: 2/3/2018 11:59:00 PM End Date: 2/5/2018 11:59:00 PM (690) Problem 15: A combination of four charges and four Gaussian surfaces are displayed in the figure. The charges have values 9,-5 nC. 92--5 nC. 93 = 19 nC, and 44 =-18 nC. 13 Otheexpertta.com 25% Part (a) what is the flux through the first surface, SI, in Nm2/C? Grade Summary Dedactiotes 0% Potl1 100% sino 789 cosn cotan() asin()| acos0 atan acotan sinh) ttempts remaining: 3 (2% per attempt) detailed view Degrees Radians lint I give up! Flints: deduction per hint. Hints remaining; Feedback: 0% deduction et feedback. · 25% Part (b) What is thie flux through the secund surface, S2-in Nm2/C? 25% Part (c) what is the flux through the third surface. S. in Nm2C? 25% Part (d) What is the flux through the forth surface, 54, in Nm-C?

Explanation / Answer

Electric flux is given by:

phi = Qnet/e0

e0 = 8.85*10^-12

A.

In first surface:

Qnet = Q1 = 5 nC

phi1 = 5*10^-9/(8.85*10^-12) = 564.97 N-m^2/C

B.

In 2nd surface:

Qnet = Q1 + Q2 = 5 nC - 5 nC = 0 nC

phi2 = 0/(8.85*10^-12) = 0 N-m^2/C

C.

In 3rd surface:

Qnet = Q2 + Q3 = -5 nC + 19 nC = 14 nC

phi3 = 14*10^-9/(8.85*10^-12) = 1581.92 N-m^2/C

D.

In 4th surface:

Qnet = Q1 + Q2 + Q3 + Q4 = 5 nC - 5 nC + 19 nC - 18 nC = 1 nC

phi4 = 1*10^-9/(8.85*10^-12) = 112.99 N-m^2/C

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