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

In summary, the left-hand sphere has a positive charge Q and the right-hand sphere has a negative charge -Q, both distributed uniformly over spherical volumes with radius R. The question asks to find the magnitude of the net electric field at a point R/2 on the x-axis. Initially, the assumption was made that the electric field would be zero, but the shell theorem states that the field inside a uniformly charged sphere is not equal to zero. After further discussion, it was determined that the smaller sphere would have 1/8 less volume, resulting in a charge of Q/8. The correct equation for the electric field for the first sphere is \frac{1}{4_{0}\epsilon\pi}\frac{Q}{\
  • #1
KayleighK
8
0

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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  • #2
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.
 
  • #3
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?
 
  • #4
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?
 
  • #5
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?
 
  • #6
Would it? How much of the original volume does the smaller sphere have?
 
  • #7
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
 
  • #8
Ok, I finally understand the problem now. Thank you so much for your help! I appreciate it =)
 

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

1. What is the formula for calculating the magnitude of an electric field on a point by 2 charged spheres?

The formula is E = (k * q * (q1/r1^2 + q2/r2^2)) / r, where E is the magnitude of the electric field, k is the Coulomb's constant, q and q1 are the charges of the spheres, r and r1 are the distances between the point and the spheres, and r2 is the distance between the two spheres.

2. How do I determine the direction of the electric field at a point between 2 charged spheres?

The direction of the electric field at a point between 2 charged spheres can be determined by using the principle of superposition. The direction of the electric field at the point is the vector sum of the individual electric fields created by each sphere. The direction will depend on the relative positions and charges of the spheres.

3. What factors affect the magnitude of the electric field on a point between 2 charged spheres?

The magnitude of the electric field on a point between 2 charged spheres is affected by the charges of the spheres, the distances between the spheres and the point, and the distance between the two spheres. The magnitude of the electric field is directly proportional to the magnitude of the charges and inversely proportional to the distance from the spheres.

4. Can the magnitude of the electric field on a point between 2 charged spheres be negative?

Yes, the magnitude of the electric field on a point between 2 charged spheres can be negative. This occurs when the directions of the individual electric fields created by the spheres are opposite and cancel each other out, resulting in a net electric field with a negative magnitude.

5. How does the magnitude of the electric field change as the distance between the spheres increases?

As the distance between the spheres increases, the magnitude of the electric field decreases. This is because the inverse square law states that the magnitude of the electric field is inversely proportional to the square of the distance between the spheres. Therefore, as the distance increases, the electric field becomes weaker.

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