Two Spheres connected by a wire.

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SUMMARY

The discussion focuses on calculating the surface charge density on two conducting spheres connected by a wire. The larger sphere, with a radius of 12 cm, has an electric field of 830 kV/m, leading to a surface charge density of 7.347 μC/m². The relationship between the charges on the spheres is established through their equal potentials when connected, using the formula kQ1/R1 = kQ2/R2. The task is to determine the charge density for the smaller sphere with a radius of 5 cm.

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Charge is placed on two conducting spheres that are very far apart and connected by a long thin wire. The radius of the smaller sphere is 5 cm and that of the larger sphere is 12 cm. The electric field at the surface of the larger sphere is 830 kV/m. Find the surface charge density on each sphere

For the 12cm sphere I used this:

[tex]E=\frac{kQ}{R^{2}}[/tex]

Solved for Q

Then divided by [tex]4\pi r^{2}[/tex], and got [tex]7.347 \frac{\mu C}{m^{2}}[/tex]

I need a little help figuring out the charge density for the 5cm radius sphere.
 
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When the two spheres are joined by a wire, their potentials are same.
So kQ1/R1 = kQ2/R2
Or (kQ1/R1^2)*R1 = kQ2/R2.
But E = (kQ1/R1^2) is given.
Find Q1 and Q2.
 

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