Need Help with Gaussian Surface Problem: Electric Flux Calculation

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The discussion revolves around calculating electric flux through a spherical Gaussian surface enclosing various charge combinations. The main equation used is φ = Q_enclosed/ε0, where Q_enclosed is the net charge within the surface. For part (a), the net charge from q1 and q2 is calculated as -0.66 μC, leading to an electric flux of approximately -7.4576 x 10^4 Nm^2/C after correcting unit conversions. The user also successfully determines that a charge of -3.05 μC would be needed to achieve zero electric flux in part (d). Overall, the conversation highlights the importance of careful unit management and understanding of electric flux calculations.
FlipStyle1308
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I just realized that I accidentally didn't post this in the homework help section...if any mods can move this, I'd appreciate it. Sorry for the trouble!

Hello. I need help solving the following problem:

Consider a spherical Gaussian surface and three charges: q1 = 1.60 μC, q2 = -2.26 μC, and q3 = 3.71 μC. Find the electric flux through the Gaussian surface if it completely encloses the following charges.

(a) only charges q1 and q2
(b) only charges q2 and q3
(c) all three charges
(d) Suppose a fourth charge, Q, is added to the situation described in part (c). Find the sign and magnitude of Q required to give zero electric flux through the surface.

I'm pretty sure that this problem involves the equation φ = q/ε0, but have no idea on how to incorporate the given values into this problem. Any help would be greatly appreciated. Thanks in advance!
 
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Maybe writing the equation this way would help: \Phi = \frac{Q_{enclosed}}{\epsilon_0}
 
Hmm...would you be able to give me an example of how to plug in my values into the equation, such as for part (a). I can try figuring out how to solve it for the rest if I get a sample with part (a). (There are no examples for this problem in both my notes and the textbook, so I have no clue on how to plug in everything.)
 
Q_enclosed is nothing but the net charge enclosed within the surface. So, in case (a), what is the net charge enclosed?
 
So for part (a) it would be 1.60μC - 2.26μC = -0.66μC? Shouldn't the answer be in Nm^2/C?
 
FlipStyle1308 said:
So for part (a) it would be 1.60μC - 2.26μC = -0.66μC? Shouldn't the answer be in Nm^2/C?
Yes, and yes.
 
Okay, so (-0.66 x 10^6 C)/(8.85 x 10^-12 C^2/N x m^2) = -7.74576 x 10^16 Nm^2/C is my answer for part (a)?
 
FlipStyle1308 said:
Okay, so (-0.66 x 10^6 C)/(8.85 x 10^-12 C^2/N x m^2) = -7.74576 x 10^16 Nm^2/C is my answer for part (a)?
The charge is given in terms of micro-coulombs, not Mega-, or million-coulombs. And it's ~ -7.4576.
 
Okay, I changed my mistake of not putting the - in when converting form micro to standard. Is the final answer -74,576.27 x 10^16 Nm^2/C?
 
  • #10
No, it's -7.4576 x 104 Nm2/C.
 
  • #11
Hmm...what did I do wrong?
 
  • #12
FlipStyle1308 said:
Hmm...what did I do wrong?
Quite simple, but if were to do the same in a test, it would get you into trouble. ;) You didn't add the powers of ten correctly.
 
  • #13
LOL okay, thanks! So I will follow the same format for (b) and (c). Any ideas on how to approach (d)?
 
  • #14
(d): It just asks for another charge that would make the flux zero, i.e. net charge Q = 0, since \epsilon_0 is a constant.
 
  • #15
So is my answer -3.05μC! Thanks so much for your help! =]
 
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  • #16
FlipStyle1308 said:
So is my answer -3.05μC! Thanks so much for your help! =]
That's correct. Sorry for the late reply.
 

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