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

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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=[tex]\frac{1}{4\pi\epsilon_{0}}[/tex] ([tex]\frac{Q}{R^{2}}[/tex]


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=[tex]\frac{1}{4\pi\epsilon_{0}}[/tex] ([tex]\frac{Q}{\frac{3}{2}R^{2}}[/tex]

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:

[tex]\frac{1}{4_{0}\epsilon\pi}[/tex][tex]\frac{Q}{\frac{1}{2}R^{2}}[/tex]

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:

[tex]\frac{1}{4_{0}\epsilon\pi}[/tex][tex]\frac{Q}{\frac{1}{2}R^{2}}[/tex]

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?
 
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 =)