# Gauss' law and spheres

1. May 15, 2015

### Brajesh kedia

Flux through a closed surface is zero due to external charge but why it is not so in case of sphere

2. May 15, 2015

### Staff: Mentor

Why do you think a sphere is treated differently than any other closed surface?

3. May 15, 2015

### Brajesh kedia

But we know flux must be zero..for external charge

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4. May 15, 2015

### Staff: Mentor

Please explain what that diagram represents.

5. May 16, 2015

### Brajesh kedia

Its a sphere of radius 3m and at distance 4m there is a charge q

6. May 16, 2015

### Staff: Mentor

I don't see any sphere in the diagram. But what about it?

If there is a sphere--or any other closed shape--and there is a charge outside the sphere, then the total flux through the surface of that sphere will be zero.

7. May 16, 2015

### Brajesh kedia

Yeah but as i shown u the above solution flux is calculater to be q/E *(1-COS@)

8. May 16, 2015

### Brajesh kedia

That round is sphere... and 4m far there is a q charge...and flux due to that charge is q/5Eo

9. May 16, 2015

### Staff: Mentor

I think what you're trying to calculate is the amount of flux from the charge through the solid angle subtended by a cone of angle θ. That's not the net flux through a closed surface.

10. May 16, 2015

### Staff: Mentor

The solid angle would be $2\pi (1 - \cos\theta)$ steradians. The total solid angle for a sphere is $4\pi$ steradians. So the fraction of the total flux from that charge going into that solid angle would be $(1 - \cos\theta)/2$.

11. May 16, 2015

### Staff: Mentor

Note that there's no "closed sphere" involved here.

12. May 17, 2015

### Brajesh kedia

I did not understand ur points..i am trying to calculate the flux through the sphere by a point charge q

13. May 17, 2015

### Staff: Mentor

Ah, I think I see what you are saying. That thing which is subtending the cone of flux (with half angle $\theta$) is a sphere. To apply Gauss' law to that sphere you must calculate the total flux through the spherical surface. What you've attempted to calculate so far is just the flux into the sphere (through the side facing the charge). You must also consider the flux going out of the sphere (through the other side of the sphere). The total flux through that spherical surface will be zero.

14. May 17, 2015

### Brajesh kedia

How to do that

15. May 17, 2015

### Staff: Mentor

Please show your work step by step. Looking at the diagram I have no idea how you came to this conclusion.

16. May 18, 2015

### Staff: Mentor

If you draw yourself a proper diagram I think you'll see that no calculation is needed to show that the net flux through the sphere is zero.

Do this. Put the charge at the origin (0,0). Then draw a sphere (a circle on your diagram) centered somewhere along the x-axis. Then draw the tangents to the circle that pass through the origin. Those will define a cone of flux from the charge at the origin. You can calculate the angle of the cone. Where the cone intersects the sphere splits the sphere into two (unequal) pieces. Clearly, whatever flux enters one side of the sphere must exit the other, so the net flux is zero.

If you wish to answer a different question, such as "how much flux from the charge hits the sphere?", then you would proceed as I outlined above. I believe that's what you were attempting to calculate yourself.

17. May 20, 2015

### Brajesh kedia

Yeah i believe net flux is zero but some are saying no..i am confused...even i knew about q=1/E not the formula which i was asked to use..the one i used

18. May 20, 2015

### Staff: Mentor

Who is saying "no" and where? Everyone in this thread seems to be saying (with remarkable patience, I might add) that the flux through the surface of a sphere when there is no charge inside of it is zero.

19. May 26, 2015

### Brajesh kedia

Do u mean that the flux i calculated is the amt of flux entering the charge.Can u please explain how formula came

20. May 26, 2015

### Staff: Mentor

I believe, as I already had stated, that you were attempting to calculate the amount of flux from the point charge hitting (entering) the sphere.

I have already done so up above. You may need to look up what a "steradian" is.