Electric Field on Two Connected Spheres

In summary, the two metallic spheres with radii of 19.1 cm and 11.5 cm have a magnitude of electric field of 3490 V/m on their surfaces. When they are connected by a long, thin metal wire, the charge is evenly distributed among the surface area. However, the potential on each point of a conductor in electrostatic equilibrium must be the same. Therefore, after connecting the spheres, the ratio of their charges is directly proportional to their radii. By equating the potential on both spheres, we can solve for the charges on each sphere, with q1=1.4E-8 C and q2=5.1E-9C.
  • #1
dfetnum
17
0

Homework Statement


Two metallic spheres have radii of 19.1 cm and 11.5 cm, respectively. The magnitude of the electric field on the surface of each sphere is 3490 V/m. The two spheres are then connected by a long, thin metal wire.

a) Determine the magnitude of the electric field on the surface of the sphere with radius 19.1 cm when they are connected.
b) Determine the magnitude of the electric field on the surface of the sphere with radius 11.5 cm when they are connected.


Homework Equations


E=kq/r^2


The Attempt at a Solution


I found the charge on each of the spheres by solving for q with the given formula. I got q1=1.4E-8 C and q2=5.1E-9C. When the spheres are connected the charge is evenly distributed among the surface area. I calculated the surface areas to be .458m^2 and .116m^2. I then found the percentage of total surface area for the first one and got .733. I used this times the sum of the total charge. I then plugged that charge into the formula. Can't seem to get the right answer.
 
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  • #2
dfetnum said:
When the spheres are connected the charge is evenly distributed among the surface area.

Perhaps surprisingly, that turns out not to be true.

When the two conductors are connected by the wire, it is as though you have one funny shaped conductor. What quantity have you studied that must have the same value at every point of a conductor when it's in electrostatic equilibrium?
 
  • #3
I don't know, electric field?
 
  • #4
No, it's not electric field. It's "electric ______" (fill in the blank)
 
  • #5
potential?
 
  • #6
Right. If an electron is free to move (as in a conductor) it will move from a point of lower potential to a point of higher potential. Since there is no movement of electrons in a conductor in electrostatic equilibrium, all points of the conductor must be at the same potential.
 
  • #7
So if I find the electric potential of both spheres before they connect, then use the proportional area, I can find the amount of potential area of one of the spheres and then multiply by the radius to find the Electric field?
 
  • #8
Don't worry about the areas of the spheres. You know the total charge. After the wire is connected, how much of that total charge should be on each sphere so that they have the same potential?
 
  • #9
I'm sorry, this is really confusing. I am just going in circles now with the numbers of the charges. I am using q1/r1=(qtotal-q1)/r2. I am getting a number larger than q total
 
  • #10
psph2.gif


Use first part of this eq. and equate V on both spheres.

You get ratio of charges in direct proportion to radii.

Solve for Q1 and Q2
 

Related to Electric Field on Two Connected Spheres

What is an electric field?

An electric field is a physical quantity that describes the strength and direction of the force that a charged particle experiences in the presence of other charged particles.

How does the electric field on two connected spheres differ from that of a single sphere?

The electric field on two connected spheres is a combination of the individual electric fields of each sphere. It is stronger in the region between the two spheres and weaker outside of them.

What factors affect the strength of the electric field on two connected spheres?

The strength of the electric field on two connected spheres depends on the charges of the spheres, their distance from each other, and the medium between them.

What is the equation for calculating the electric field on two connected spheres?

The equation for calculating the electric field on two connected spheres is E = kQ/(r1^2) + kQ/(r2^2), where k is the Coulomb's constant, Q is the charge on each sphere, and r1 and r2 are the distances from the spheres.

How can the direction of the electric field on two connected spheres be determined?

The direction of the electric field on two connected spheres is determined by the direction that a positive test charge would move in the presence of the two spheres. It always points away from positively charged spheres and towards negatively charged spheres.

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