Two Spheres connected by a wire.

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The discussion focuses on calculating the surface charge density of 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 calculated surface charge density of approximately 7.347 µC/m². To find the charge density for the smaller 5 cm sphere, the relationship between their potentials is utilized, indicating that the potentials must be equal when connected. The equations kQ1/R1 = kQ2/R2 and E = (kQ1/R1²) are key to solving for the charges Q1 and Q2. The discussion emphasizes the importance of understanding electric fields and potential in connected conductive systems.
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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:

E=\frac{kQ}{R^{2}}

Solved for Q

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

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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