3
$\begingroup$

We have a plate capacitor:

The picture shows this plate capacitor where the space between the electrodes is filled with a dielectric plate with dielectric constant $E_r = 8$:

  • Determine the length of $h$ to which should extract dielectric plate, in order to reduce the capacity of capacitor by 4 times
  • What will be the percentage decrease in capacitance if we pull the plate to a length of $a/5$?
$\endgroup$
1
  • 2
    $\begingroup$ Welcome to Engineering! This looks like a homework question. In order for such questions to be answered in this site, we need you to add details describing the precise problem you're having. What have you tried to solve this yourself? Please edit your question to include this information. $\endgroup$
    – Wasabi
    Aug 16 '16 at 13:35
2
$\begingroup$

Since we know from basic:

C = ε.A/H

Which is in our case:

C = ε.(b.a)/d

In order to reduce the capacity of capacitor for 4 times

  1. b = b/4 (Divide length by 4)
  2. a = a/4 (Divide width by 4)
  3. d = d.4 (Multiply height by 4)
  4. or play with constant ε (find some material which has ε' as ε/4)

For what percentage will decrease the capacitance, if we divide the plate's width by 5 ?

For that you can use the same equation:

C' = ε.(b.(a/5))/d

Then you can get: (C'/C)*100 = %20.

This means that it gets smaller as the %20 of the original value.


C (Capacity) = ε (Vacuum permittivity) . A (Area) / H (Height)

For more information, please take a look at Capacitor Wiki.

$\endgroup$
0
$\begingroup$

The plate system above can effectively thought as two parallel capacitors.

Both will have b length and d thickness. Then leftside one will have h width and Er=1 as electric constant, while the rightside one will span by a-h and score Er=4.

Now, simply enough, in the usual approximation for flat plate capacitors (i.e. give up edges and consider inner field constant and outer one naught )

$$ C=C_\text{L}+C_\text{R}=\epsilon_0\frac{h\,b}{d}+\epsilon_0 \epsilon_\text{r}\frac{(a-h)b}{d} $$

Which by some algebra can be rearranged as

$$ C=\epsilon_0\frac{a\,b}{d}\left(\epsilon_\text{r}-(\epsilon_\text{r}-1)\frac{h}{a}\right ) $$

This can be read as one a by b capacitor with average relative electric constant linearly changing from Er to 1 as h/a ratio spans 0 to 1

$\endgroup$

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