Question

The unit Farad is not equivalent to
A) $C{V^2}$
B) $\dfrac{J}{{{V^2}}}$
C) $\dfrac{{{Q^2}}}{J}$
D) $\dfrac{Q}{V}$

Answer 1

Hint: Farad is the unit of capacitance. Capacitance is denoted by $C$.
Let the charge be $Q$, $C$ be the capacitance, $V$ be the potential difference and $J$ be the potential energy stored in the capacitor.
By using the expressions-
$Q = CV$
$J = \dfrac{1}{2}C{V^2}$ and
$J = \dfrac{{{Q^2}}}{{2C}}$
Now, do the transpositions and find the expressions to which $C$ is equivalent.

Complete step by step answer:
The two-terminal electrical device that has the capability to store energy in the form of an electrical charge is called a Capacitor. The capacity of a capacitor to store energy in the form of charge is called capacitance.
The insulating material which is used to fill the gap between the conductors with the vacuum is called dielectric.
The SI unit of the capacitance is Farad which is denoted by $F.$
The SI unit of the capacitance is equivalent to many other units. The equivalent units to farad are shown below:
Let the charge be $Q$, $C$ be the capacitance, $V$ be the potential difference and $J$ be the potential energy stored in the capacitor.
We know that,
$
  Q = CV \\
  \Rightarrow C = \dfrac{Q}{V} \\
 $
Expressing in dimensions. Then,
$\Rightarrow \left[ F \right] = \left[ {\dfrac{Q}{V}} \right] \cdots (1)$
Also, $J = \dfrac{1}{2}C{V^2}$
Therefore, by transposition, we get
$\Rightarrow C = \dfrac{{2J}}{{{V^2}}}$
In dimensional formula
then, $\left[ F \right] = \left[ {\dfrac{J}{{{V^2}}}} \right] \cdots (2)$
And, we also know that,
$\Rightarrow J = \dfrac{{{Q^2}}}{{2C}}$
By transposition, we get
$\Rightarrow C = \dfrac{{{Q^2}}}{{2J}}$
In dimensional expression
then, $\left[ F \right] = \left[ {\dfrac{{{Q^2}}}{J}} \right] \cdots (3)$
Therefore, from equations $(1),(2)$ and $(3)$, we get
$\Rightarrow F = \dfrac{Q}{V} = \dfrac{J}{{{V^2}}} = \dfrac{{{Q^2}}}{J}$

Therefore, Farad is not equivalent to $C{V^2}$. Hence, option (A) is the correct answer.

Note:
One farad can be defined as the capacitance across which, when charged with one coulomb of charge then, there is a potential difference of one volt.
The relationship between capacitance, charge, and the potential difference is linear.