1
\$\begingroup\$

enter image description here

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<Q_{1}=-Q_{2} \leftrightarrow \left| Q_{1} \right| =\left| Q_{2} \right| \$

I can determine \$Q=Q_{1}\$ but how about it is not guaranteed of \$0<Q_{1}=-Q_{2} \leftrightarrow \left| Q_{1} \right| =\left| Q_{2} \right| \$

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.

\$\endgroup\$
2
  • \$\begingroup\$ 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. \$\endgroup\$
    – andrea
    Jun 19, 2021 at 6:43
  • \$\begingroup\$ From gauss law , outer cylinder has charges -Q1 on inner surface and Q2+Q1 on outer surface , so you can safely assume Q=Q1 \$\endgroup\$
    – user215805
    Jun 19, 2021 at 7:00

1 Answer 1

0
\$\begingroup\$

The capacitor in a static electricity apparatus is sometimes just a sphere on a stick. The capacitance of such an item is calculated with the assumption of a very-large-container around the sphere, which is grounded. This 'ground at infinity' means that a second electrode is assumed to exist, with access to whatever charge it 'needs',and that means the second electrode carries equal and opposite charge to the first. Exact dimensions of an outer container, if it is sufficiently large, are irrelevant (the E-fields fall off with distance, all the capacitive field energy is near the sphere on the stick).

In circuits (not static electricity) a capacitor has two terminals, but the same assumption can be made, that the (outer, usually) electrode is grounded, i.e. has access to an inexhaustible charge reservoir, and thus will always take on an equal and opposite charge to the inner.

Since the imbalance of charge on the two plates does not change the capacitance (which is a dimension-geometry-and-materials dependent number), you can safely assume that the outer holds charge -Q1, if that helps the calculation. Then the 'Q' of your formula is the charge on the inner electrode, Q1.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.