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Question: A capacitor consists of two metal plates each 10cm by 20cm, they are separated by a 2mm thick insula...

A capacitor consists of two metal plates each 10cm by 20cm, they are separated by a 2mm thick insulator with dielectric constant 4.1 and dielectric strength 6.0107V/m. What is the capacitance in pF (10−12{{10}^{-12}}F)?
(A). 75
(B). 100
(C). 240
(D). 360

Explanation

Solution

Hint: Capacitor is two parallel plates separated by small distance. Capacitance of the capacitor is defined as the ratio of electrical charge on each conductor to the potential difference between them. Capacitance depends on shape, size, and geometrical placing of conductors and the medium in between them.

Complete step by step answer:
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Consider a parallel plate capacitor filled with insulator of dielectric constant k with charge +Q and -Q as shown below:

Capacitance of the parallel plate capacitor is given by the formula
C=kε0AdfaradC=k{{\varepsilon }_{0}}\dfrac{A}{d}\text{farad} ……… (1)
Where,
k – dielectric constant, it is due to the introduction of dielectric medium between the plates of the capacitor.
ε0{{\varepsilon }_{0}}- permittivity of free space, which is the measure of how much electric field lines a medium allows to pass through.
A - area of capacitor plate
d - distance between the two plates
Now, consider above question, here they have given that-
k = 4.1
ε0=8.854×10−12Nm2C−2{{\varepsilon }_{0}}=8.854\times {{10}^{-12}}N{{m}^{2}}{{C}^{-2}}
A=(10cm×20cm)=200×10−4m2A=(10cm\times 20cm)=200\times {{10}^{-4}}{{m}^{2}}
d=2mm=2×10−3md=2mm=2\times {{10}^{-3}}m
Substituting these values in equation (1), we get,
C=kε0Ad faradC=k{{\varepsilon }_{0}}\dfrac{A}{d}\text{ farad}
=4.1×8.85×10−12×2×10−22×10−3  \begin{aligned} & =\dfrac{4.1\times 8.85\times {{10}^{-12}}\times 2\times {{10}^{-2}}}{2\times {{10}^{-3}}} \\\ & \\\ \end{aligned}
On calculating we get,
=36.3×10−11farad=36.3\times {{10}^{-11}}\text{farad}
∴C=363×10−12F≃360pF\therefore C=363\times {{10}^{-12}}F\simeq 360pF
Thus, the correct option is D.

Note: Remember the formula of capacitance to solve this type of question.
Students usually make mistakes while substituting for k as dielectric strength value is also given in the question, but k is a dielectric constant not dielectric strength. To remember it easily, notice that the dielectric constant has no unit.
Dielectric strength is maximum electric field strength above which an insulating material begins to break down and its unit is V/m.
When your answer is not as exactly as the option given in the question then approximate it to the nearest answer.
Always keep the power value in answer as asked in the question and the unit should be proper.
And one more mistake student make is while calculating area , consider above values only when we multiply 10cm10cm and 20cm20cm then answer is 200(cm)2=200×10−4200{{\left( cm \right)}^{2}}=200\times {{10}^{-4}} but students make mistake by writing it 200×10−2m2200\times {{10}^{-2}}{{m}^{2}}.