Question

A variable air capacitor has 11 movable plates and 12 stationary plates, the area of each plate is $0.0015{{m}^{2}}$ and separation between opposite plates is 0.001m. The maximum capacitance of the capacitor is
A. 292.2F
B. 292.2mF
C. 292μF
D. 292pF

Answer 1

From the question we get that the given capacitor is a multi-plate capacitor with air as the dielectric medium. You could now recall the expression for capacitance of the capacitor with n plates and then substitute the sum of the movable and stationary plate in place of n. Also, ${{\varepsilon }_{r}}$ here is that of air=1.0006 and other quantities like area and distance of separation between plates are directly given in the question.
Formula used:
Expression for capacitance of a parallel plate capacitor with n plates,
$C=\dfrac{A{{\varepsilon }_{0}}{{\varepsilon }_{r}}\left( n-1 \right)}{d}$

Complete answer:
We know that a capacitor is more like a small rechargeable battery storing energy in the form of electrical charge producing a potential difference across its plates. Basically a capacitor consists of two conductors separated by an insulator. This insulating layer is what we call a dielectric. The capacitance of capacitor is given by,
$C=\dfrac{Q}{V}$
It should be noted that C is independent of Q and V but depends on geometrical configuration, that is, shape, size, and separation of the system of conductors. Its SI unit is farad.
Now let us discuss a parallel plate capacitor. Here, the capacitance is directly proportional to the area (A) of the conductive plates and is inversely proportional to the distance of separation (d).
$C\propto \dfrac{A}{d}$
$C=\varepsilon \dfrac{A}{d}$
Where, ε is the absolute permittivity of the dielectric material. We also have permittivity of vacuum/permittivity of free space ${{\varepsilon }_{0}}=8.84\times {{10}^{-12}}F{{m}^{-1}}$. Also when vacuum of free space is replaced by some other material, complex permittivity comes into picture which is given by,
$\varepsilon ={{\varepsilon }_{0}}\times {{\varepsilon }_{r}}$
Thereby, we get the final expression for capacitance as,
$C=\dfrac{A{{\varepsilon }_{0}}{{\varepsilon }_{r}}}{d}$ …………………….. (1)
Where, ‘${{\varepsilon }_{r}}$’ is the permittivity of the medium.

We also have the scope of increasing the capacitance by interleaving more plates together within a single capacitor body.
We can rewrite equation (1) as,
$C=\dfrac{A{{\varepsilon }_{0}}{{\varepsilon }_{r}}\left( n-1 \right)}{d}$………………… (2)
Where n is the total number of plates in a capacitor. Substituting n=2 in (2) we can end up in (1).
In the question we are given a multi-plate variable air capacitor, in which air forms the dielectric, which means,
${{\varepsilon }_{r}}=1.0006$ …………………. (3)
And the given capacitor has 11 movable and 12 stationary plates, that is, it has a total of 23 plates. So,
$n=23$ ………………………. (4)
Area of each plate is,
$A=0.0015{{m}^{2}}$…………………….. (5)
Separation between opposite plates is,
$d=0.001m$ ………………………………. (6)
Substituting (3), (4), (5) and (6) in (2), we get the capacitance of the given capacitor as,
$C=\dfrac{0.0015\times 8.84\times {{10}^{-12}}\times 1.0006\left( 23-1 \right)}{0.001}$
$C=291.895\times {{10}^{-12}}F\sim 292pF$
Therefore, the maximum capacitance of the given capacitor is 292pF.

Hence, the answer to the given question is option D.

Note:
The equation (2) is based on certain logic. Consider the parallel plate capacitor with two plates. Though it has two plates, only one side of each plate is in contact with the dielectric placed in between them when the other side of these plates forms the outside of the capacitor. This is also true for multi-plate capacitors as only one side of the plates at the extreme ends is in contact with the dielectric. Hence, the term (n-1) comes into picture.