Solveeit Logo
Login

Question

Question: \- The formula for the capacity of condenser is given by \(C = \dfrac{A}{d}\) When \(A\) is the area...

- The formula for the capacity of condenser is given by C=AdC = \dfrac{A}{d} When AA is the area of each plate and dd is the distance between the plates. Then the dimensions of missing quantity is:
A)∈0=M−1L−3T4A2{ \in _0} = {M^{ - 1}}{L^{ - 3}}{T^4}{A^2}
B)∈0=M1L3T−4A−2{ \in _0} = {M^1}{L^3}{T^{ - 4}}{A^{ - 2}}
C)∈0=M−1L3T4A−2{ \in _0} = {M^{ - 1}}{L^3}{T^4}{A^{ - 2}}
D)∈0=M−1L−2T4A2{ \in _{_0}} = {M^{ - 1}}{L^{ - 2}}{T^4}{A^2}

Explanation

Solution

Condenser capacity defined as the ability of heat transfer from hot gas (vapours) to the surrounding condensing medium Liquid condensing medium (water) is more effective than the gaseous condensing than the gaseous condensing medium (air).

Complete step by step answer:
Whenever an electric charge is deposited on a 44 conductor, its potential increases the deposited charge spreads over its surface for any conductor, the electric potential (v)(v) is directly proportional to the charge store (Q)(Q).
Hence QQ α\alpha vv
Q=cvQ = cv
Where ccis a constant known as capacity of the conductor.
In this given problem we have to find the dimension of missing quantity.
Let missing quantity be ∈0{ \in _0}
C=∈0AdC = \dfrac{{{ \in _0}A}}{d}
As per the given details
Now,
Applying the dimensional formula per primitively of free space
M−1L−2T4A2=∈0L2L{M^{ - 1}}{L^{ - 2}}{T^4}{A^2} = { \in ^0}\dfrac{{{L^2}}}{L}
⇒∈0=M−1L−3T4A2\Rightarrow { \in _0} = {M^{ - 1}}{L^{ - 3}}{T^4}{A^2}

So the correct option is (A) ∈0=M−1L−3T4A2{ \in _0} = {M^{ - 1}}{L^{ - 3}}{T^4}{A^2}

Additional Information:
The capacity of a conductor (c)(c)is defined as the amount of charge required to make the potential 11 unit (1volt)(1volt)
Capacity of a conductor is given by
c=Qvc = \dfrac{Q}{v}
Let v=1vv = 1v, then c=Qc = Q
The capacity of conductor (c)(c)is defined as the amount of charge required to make the potential 11 unit (1volt)(1volt)
SI unit of capacity is farad (F)(F)
We have c=c = c=Qvc = \dfrac{Q}{v}
SI unit of capacity ==SI unit of electric charge // SI unit of potential difference.
Therefore, 1 farad = 1 coulomb //1 volt
The capacity of conductor is 11 farad if a charge of 11 Coulomb raises its potential by 11 volt.
The farad is a very large unit hence smaller partial units are used.
Smaller units are,
11 micro farad (1μF)=10−6F(1\mu F) = {10^{ - 6}}F
1 nanofarad (1ηF)=10−9F(1\eta F) = {10^{ - 9}}F
11 picofarad (1ρF)=10−12F4(1\rho F) = {10^{ - 12}}F4

Note:
we can choose a condenser capacity by applying following steps:
Step1: Select a solvent having high volatility
Step2: Calculate the maximum boil-up of the selected solvent.
Step3: Calculate the condenser capacity for that boil-up