Other unit for the quantity having the units `(C^(2))/(Nm^(2))` is

Options

A. farad
B. `("farad")/(m^(2))`
C. `("farad")/(m)`
D. `(m)/("farad")`

Answer

To find another unit for the quantity having the units \((C^{2})/(Nm^{2})\), we can follow these steps:

### Step 1: Rewrite the given units
We start with the given units:
\[ \frac{C^{2}}{Nm^{2}} \]
This can be rewritten as:
\[ \frac{C}{N \cdot m^{2}} \cdot C \]
This shows that we have Coulomb squared divided by Newton times meter squared.

### Step 2: Substitute the unit of Newton
We know that:
\[ 1 \, \text{Newton} = 1 \, \text{kg} \cdot \text{m/s}^{2} \]
Substituting this into our expression gives:
\[ \frac{C^{2}}{(kg \cdot m/s^{2}) \cdot m^{2}} = \frac{C^{2} \cdot s^{2}}{kg \cdot m^{3}} \]

### Step 3: Relate to electric potential
We know that electric potential \(V\) can be defined as:
\[ V = \frac{W}{Q} \]
Where \(W\) is work done (in Joules) and \(Q\) is charge (in Coulombs). The unit of work (Joule) can also be expressed as:
\[ 1 \, \text{Joule} = 1 \, \text{Newton} \cdot \text{meter} = 1 \, \frac{kg \cdot m^{2}}{s^{2}} \]
Thus, we can express \(V\) in terms of units:
\[ V = \frac{Nm}{C} \]

### Step 4: Substitute \(V\) back into the expression
We can rearrange the expression for \(V\) to isolate \(C/V\):
\[ \frac{C}{V} = \frac{C \cdot C}{Nm} = \frac{C^{2}}{Nm} \]
Thus, we can express our original units as:
\[ \frac{C^{2}}{Nm^{2}} = \frac{C}{V} \cdot \frac{1}{m} \]

### Step 5: Relate to capacitance
We know that capacitance \(C\) is defined as:
\[ C = \frac{Q}{V} \]
The unit of capacitance is Farad (F). Therefore, we can express:
\[ \frac{C^{2}}{Nm^{2}} = \frac{F}{m} \]

### Conclusion
Thus, we conclude that the other unit for the quantity having the units \((C^{2})/(Nm^{2})\) is:
\[ \text{Farad per meter} \, (F/m) \]