Electric Field on Two Connected Spheres

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When two metallic spheres are connected by a wire, the electric potential across both spheres must equalize, leading to a redistribution of charge. The initial charges on the spheres are calculated as q1=1.4E-8 C and q2=5.1E-9 C, but these need to be adjusted based on their radii to maintain equal potential. The electric field on the surface of each sphere can be determined using the formula E=kq/r^2, but the total charge must be divided according to the ratio of their radii. The discussion emphasizes that the electric field is not uniform across the spheres when connected, and the correct approach involves equating the potentials to find the new charge distribution. Understanding this concept is crucial for solving the problem accurately.
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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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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?
 
I don't know, electric field?
 
No, it's not electric field. It's "electric ______" (fill in the blank)
 
potential?
 
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.
 
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?
 
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?
 
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
 
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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
 
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