Two small insulating spheres with radius

In summary, Two insulating spheres with net charges of -2.20 \mu C and 4.05 \mu C are separated by a center-to-center distance of 0.510 m. The charge is uniformly distributed within the volume of each sphere. To calculate the magnitude of the electric field at the midway point due to the sphere with charge -2.20 \mu C only, you can use the formula E(sphere) = Q/ 4pi(distance)^2, where the distance is half of the center-to-center distance. It is not necessary to use the radius of the sphere in this calculation. The issue with the radius may have been due to not converting the charge from micro coulombs to coulombs
  • #1
viperassasin
11
0

Homework Statement



Two small insulating spheres with radius 5.50×10−2 m are separated by a large center-to-center distance of 0.510 m. One sphere is negatively charged, with net charge -2.20 \mu C, and the other sphere is positively charged, with net charge 4.05 \mu C. The charge is uniformly distributed within the volume of each sphere.

Calculate E_1, the magnitude of the electric field at the midway point due to the sphere with charge -2.20 \mu C only.

Homework Equations





The Attempt at a Solution


I tried to use q/4pi(espolon not)r^2 but I am not sure my radius is right because I have used half the distance for the distance and the radius of the sphere for the distance.
 
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  • #2
E(sphere) = Q/ 4pi(distance)^2
Notice that this the same as a point charge. You do not need to worry about the radius of the sphere for this part of the question.
 
  • #3
thanks

my problem was that I wasn't changing my units from micro coulombs to coulombs.
 

Related to Two small insulating spheres with radius

1. What is the formula for the electric potential of two small insulating spheres with radius?

The formula for the electric potential of two small insulating spheres with radius is V = kQ/r, where k is the Coulomb's constant, Q is the charge of the sphere, and r is the distance between the two spheres.

2. How do the charges on the two spheres affect the electric potential?

The charges on the two spheres affect the electric potential by creating an electric field between them. The strength of the electric field depends on the magnitude of the charges and the distance between the spheres. As the charges increase, the electric potential also increases.

3. What is the significance of the radius of the insulating spheres in this scenario?

The radius of the insulating spheres is significant because it determines the distance between the two spheres and affects the electric potential. As the radius decreases, the distance between the spheres decreases, resulting in a stronger electric potential.

4. How does the distance between the two spheres affect the electric potential?

The distance between the two spheres is directly proportional to the electric potential. As the distance increases, the electric potential decreases. This is because the electric field weakens as the distance increases.

5. Can the electric potential of two small insulating spheres be negative?

Yes, the electric potential of two small insulating spheres can be negative. This occurs when the charges on the spheres have opposite signs. In this case, the electric potential is negative because the electric field between the spheres is directed from the positive charge to the negative charge.

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