Question

with n parallel plate each of same area A and separate by equal distance d with air or vacuum between the plate what is the effective capacity?​

Answer 1

As per the description given in the question, it can be inferred that they are capacitors in a series.

Now, it is known that, capacitors in a series   act like resistors do in a parallel  structure.

⇒ $\frac{1}{Ceff} =$ ∑$\frac{1}{Ci}$

in which, Ci = capacitance of the capacitor

Now, we know,

$Ci =$ ε $\frac{A}{d}$

∴  ∑$\frac{1}{Ci}$ for $n_{2}$ = (2-1) = 1 capacitor

Similarly,  ∑$\frac{1}{Ci}$ for $n_{3}$ = (3-1) = 2 capacitors

Thus, ∑$\frac{1}{Ci}$  for (n = n) = (n-1) capacitors

∴ $Ceff$ =  ε $\frac{A}{d} (n-1)$

Ans) $Ceff$ =  ε $\frac{A}{d} (n-1)$

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Answer 2

Given:

n parallel plate each of same area A and separate by equal distance d with air or vacuum between the plate.

To find:

The effective capacitance ?

Calculation:

When there are "n" plates parallel to one another, we can easily understand that their is (n-1) capacitors attached end to end.

• This means that (n-1) capacitors are attached in series combination.

Now, we know that in series combination the net potential difference supplied by the cell is equal to the the sum of the individual potential drops along each of the capacitor.

$\therefore \: V_{net} = V_{1} + V_{2} + ..... + V_{(n - 1)}$

$\implies \: \dfrac{q}{C_{net}} = \dfrac{q}{C_{1}} + \dfrac{q}{C_{2}} + ..... + \dfrac{q}{C_{(n - 1)}}$

$\implies \: \dfrac{1}{C_{net}} = \dfrac{1}{C_{1}} + \dfrac{1}{C_{2}} + ..... + \dfrac{1}{C_{(n - 1)}}$

Now, as per the question, we know that all the capacitors have same capacitance = C (let):

$\implies \: \dfrac{1}{C_{net}} = \dfrac{1}{C} + \dfrac{1}{C} + ..... (n - 1) \: times$

$\implies \: \dfrac{1}{C_{net}} = \dfrac{(n - 1)}{C}$

$\implies \: C_{net} = \dfrac{C}{(n -1 )}$

$\implies \: C_{net} = \dfrac{ (\frac{q}{V}) }{(n -1 )}$

$\implies \: C_{net} = \dfrac{ (\frac{q}{E \times d}) }{(n -1 )}$

$\implies \: C_{net} = \dfrac{ (\dfrac{q}{ \frac{ \sigma}{ \epsilon_{0}} \times d}) }{(n -1 )}$

$\implies \: C_{net} = \dfrac{ (\dfrac{q}{ \frac{q}{ A\epsilon_{0}} \times d}) }{(n -1 )}$

$\implies \: C_{net} = \dfrac{ (\dfrac{1}{ \frac{1}{ A\epsilon_{0}} \times d}) }{(n -1 )}$

$\implies \: C_{net} = \dfrac{A \epsilon_{0} }{d(n - 1)}$

So, final answer is:

$\boxed{ \bold{ \: C_{net} = \dfrac{A \epsilon_{0} }{d(n - 1)} }}$