Two Spheres connected by a wire.

[BustedBreaks]

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.

 

[rl.bhat]

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.

 

[tomcue]

Based on the given information, we can calculate the electric field at the surface of the smaller sphere using the same equation:

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

Where Q is the total charge on the smaller sphere and R is the radius of the smaller sphere.

Substituting the values, we get:

830 kV/m = \frac{kQ}{(5 cm)^2}

Solving for Q, we get:

Q = 1.037 x 10^{-10} C

To find the surface charge density, we can divide this charge by the surface area of the smaller sphere:

\sigma = \frac{Q}{4\pi (5 cm)^2}

Plugging in the values, we get:

\sigma = 1.037 \frac{\mu C}{m^{2}}

Therefore, the surface charge density on the smaller sphere is 1.037 \frac{\mu C}{m^{2}}.