Question

A certain capacitor has rectangular plates 59 cm by 39 cm, and the gap width is 0.24 mm. if the gap is filled with a material whose dielectric constant is 3.3, what is the capacitance?

Answer 1

The capacitance of the certain capacitor (C) = 28.02 nF.

How can we calculate the value of capacitance?

To calculate the value of capacitance of the certain capacitor we are using the formula here is,

$C= K \times \frac{\epsilon_0 A}{d}$

Here we are given,

K= Dielectric constant of a material by which the capacitor gap is filled with = 3.3

$\epsilon_0$= The permittivity of free space = $8.854 \times 10^{-12} C^2/Nm^2$

d= The width of the gap = 0.24 mm = 0.00024m.

A= The area of the plates of the capacitor.

As we know,

A = l × b

Here we are given,

l = length of the plate = 59 cm.

b = breath of the plate = 39 cm.

So according to the formula, area of the plate is,

A = l₁ × l₂

   = 59 × 39

   = 2301 cm²

   = 0.2301 m²

So, The area of the plates of the capacitor (A)= 0.2301$m^2$

We have to find the capacitance of a certain capacitor, C.

Now we put the known values in the above equation, we can get,

$C= K \times \frac{\epsilon_0 A}{d}$

Or, $C= 3.3 \times \frac{8.854 \times 10^{-12} \times 0.2301}{0.00024}$

Or, $C= 2.802\times 10^{-8} F$

Or,$C= 28.02\times 10^{-9} F$

Or,$C= 28.02 nF$

From the above calculation we can conclude that, the capacitance of the certain capacitor (C) is 28.02 nF.

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