Magnitude of an Electric field on a point by 2 charged spheres

AI Thread Summary
The discussion revolves around calculating the net electric field at a point R/2 on the x-axis due to two uniformly charged spheres, one positive and one negative. Initially, there was confusion about the electric field being zero inside the first sphere, but it was clarified that the field is not zero since the sphere is not a conductor. The correct approach involves applying the shell theorem, which states that only the charge within a radius of R/2 contributes to the electric field at that point. The participants concluded that the charge contributing to the field at R/2 is effectively reduced, leading to a better understanding of the problem. The discussion highlights the importance of recognizing how charge distribution affects electric fields within non-conductive spheres.
KayleighK
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Homework Statement



The left-hand sphere has a positive charge Q and the right-hand sphere has a negative charge -Q . Charge is distibuted uniformly over each of two spherical volumes with radius R. One sphere of charge is centered at the origin and the other at x=2R .
Find the magnitude of the net electric field at the point R/2 on the x-axis



Homework Equations



E=\frac{1}{4\pi\epsilon_{0}} (\frac{Q}{R^{2}}


The Attempt at a Solution



Since the point is located within the first sphere, I thought the electric field would be zero.
Then I typed in:

E=\frac{1}{4\pi\epsilon_{0}} (\frac{Q}{\frac{3}{2}R^{2}}

but it said the answer was wrong. Can anyone please help?
 
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Since the point is located within the first sphere, I thought the electric field would be zero.

But the sphere is not a conductor. So the field inside a uniformly charge sphere the is not equal to zero.
 
rl.bhat said:
Since the point is located within the first sphere, I thought the electric field would be zero.

But the sphere is not a conductor. So the field inside a uniformly charge sphere the is not equal to zero.

Oh ok, so then the E field for the first sphere would be:

\frac{1}{4_{0}\epsilon\pi}\frac{Q}{\frac{1}{2}R^{2}}

And then since both fields of the spheres are pointing the same direction I will just add the E fields of the first sphere with the second sphere?
 
KayleighK said:
Oh ok, so then the E field for the first sphere would be:

\frac{1}{4_{0}\epsilon\pi}\frac{Q}{\frac{1}{2}R^{2}}

Why? The shell theorem states that the part of the sphere from r=1/2R to R creates no net electric field. The sphere "under" that shell has radius 1/2R, so what charge must it have?
 
ideasrule said:
Why? The shell theorem states that the part of the sphere from r=1/2R to R creates no net electric field. The sphere "under" that shell has radius 1/2R, so what charge must it have?

Would it have half the charge it originally had? Q/2?
 
Would it? How much of the original volume does the smaller sphere have?
 
ideasrule said:
Would it? How much of the original volume does the smaller sphere have?

It would have 1/8 less volume...so then Q/8
 
Ok, I finally understand the problem now. Thank you so much for your help! I appreciate it =)
 
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