It's easy. A capacitor **doesn't** store charge, it stores **energy**. The net charge in a complete capacitor (rather than considering a single plate or the insulator) never changes. An increase of negative charge on one plate is exactly balanced by a decrease in negative charge on the other plate. Therefore, as current enters one terminal an equal current must leave the other terminal.

This is a nice question...

Let's start with your last statement. It must be

$$\lambda_c=-(\lambda_a+\lambda_b)$$

because cylinder C has to cancel out the charges of cylinders A and B.

Now, think about the fields generated by the three charges. It looks like this:

Remember, that the charge on a conducting, hollow body does not generate any E-field inside.

But the E-field of a charge inside a hollow body does not end at the surface, it expands to infinity.

This also is a result of Gauß' law, as it simply says that *field times (closed) surface is charge inside that surface*.

The superposition of all fields is the grey curve, and now you see that

$$V_b=\int_b^c\frac{(\lambda_a+\lambda_b)}{2\pi\varepsilon_0r}\,dr$$

and

$$V_a=\int_a^b\frac{\lambda_a}{2\pi\varepsilon_0r}\,dr+V_b$$

With given potential(difference)s, you can calculate \$\lambda_a\$ and \$\lambda_b\$.

## Best Answer

The electrostatic field is zero AFTER equilibrium is established (after the internal E-field cancels the external E-field.)

Remember the equation that describes the introduction of charges at some interior point of a given material:

What it means is that any charge introduced at the interior of the conductor will decay immediately and appear on the the surface.

Stated in other way, if some charges are introduced in the interior of a conductor, the charges will move to the surface and redistribute themselves quickly in such a manner that the field inside the conductor vanishes.