# As amounts of the 2 charges are given only by symbols , how do I determine the amount of charge of $Q=CV~$ of cylindrical capacitor?

The cylinder exists with the height $$\d\$$ and the radius $$\a\$$

The cylindrical shell surrounds that cylinder with the cocentric radius $$\b\$$

The space between of it has been filled with the dielectric of $$\ \epsilon_{} \$$

$$\Q_{1},Q_{2} \$$ are given to the inner,outer conductors respectively.

I want to calculate the capacitance of this capacitor.

First things to first, the electric field inside the dielectric is easily obtained by

$$\left( 2\pi r \cdot d \right) E_{r} = \frac{ Q_{1} }{ \epsilon_{} }$$

$$E_{r} = \frac{ Q_{1} }{ 2\pi rd \epsilon_{} }$$

To find out the voltage between the conductors,

$$V= -\int_{b }^{ a} \frac{ Q_{1} }{ 2\pi rd \epsilon_{} } \,dr$$

$$= \int_{a }^{ b} \frac{ Q_{1} }{ 2\pi rd \epsilon_{} } \,dr$$

$$= \frac{ Q_{1} }{ 2\pi d \epsilon_{} } \int_{a }^{ b} \frac{ 1 }{ r} \,dr$$

$$= \frac{ Q_{1} }{ 2\pi d \epsilon_{} } \ln\left( b/a \right)$$

The problem begins from here.

I attempted to use the general formula $$\CV=Q\$$

$$C=\frac{ Q }{ V }$$

How the value of $$\Q\$$ is determined?

As the distributions of charges are one of the typical patterns like $$\0

I can determine $$\Q=Q_{1}\$$ but how about it is not guaranteed of $$\0

Or can I assume $$\ \left| Q_{1} \right| =\left| Q_{2} \right|\$$ forcefully?

By the way I assumed that the any electric field is vertical against the surface of the flank of the inner cylinder. Is it correct?

The inner conductor is given $$\Q_{1}\$$ but the distribution of the charges is undefined.

• I do not see the problem: in your formulas you have V as a function of Q, so if you put in the fraction Q and V, then Q will rule out and disappear whatever its value. The only assumption is that Q is the same but with opposite sign on the two surfaces of the cylinder. -- and yes, the electric field is vertical, except at the two ends where you have some deviation from it. Commented Jun 19, 2021 at 6:43
• From gauss law , outer cylinder has charges -Q1 on inner surface and Q2+Q1 on outer surface , so you can safely assume Q=Q1 Commented Jun 19, 2021 at 7:00