The dimensional formula for permittivity of free space `(epsilon_0)` in the equation `F ==1/(4piepsilon_0),(q_1q_2)/(r^2)` , where symbols have usual meaning is

Options

A. `[M^1L^3A^(-2)T^(-4)]`
B. `[M^(-1)L^(-3)T^4A^2]`
C. `[M^(-1)L^(-3)A^(-2)T^(-4)]`
D. `[M^(1)L^(3)T^2A^(-4)]`

Answer

To find the dimensional formula for permittivity of free space \((\epsilon_0)\) in the equation

\[ F = \frac{1}{4\pi\epsilon_0} \cdot \frac{q_1 q_2}{r^2} \]

we will follow these steps:

### Step 1: Rearranging the Equation
We start by rearranging the equation to isolate \(\epsilon_0\):

\[ \epsilon_0 = \frac{1}{4\pi} \cdot \frac{q_1 q_2}{F \cdot r^2} \]

### Step 2: Identifying Dimensions
Next, we need to identify the dimensions of each component in the equation:

1. **Force (F)**: The dimensional formula for force is given by:
\[ [F] = MLT^{-2} \]
where \(M\) is mass, \(L\) is length, and \(T\) is time.

2. **Distance (r)**: The dimensional formula for distance is:
\[ [r] = L \]
Therefore, the dimensional formula for \(r^2\) is:
\[ [r^2] = L^2 \]

3. **Charge (q)**: The dimensional formula for electric charge is:
\[ [q] = IT \]
where \(I\) is current (in amperes) and \(T\) is time. Therefore, for two charges \(q_1\) and \(q_2\):
\[ [q_1 q_2] = (IT)(IT) = I^2T^2 \]

### Step 3: Substituting Dimensions into the Equation
Now we can substitute these dimensions back into the equation for \(\epsilon_0\):

\[ [\epsilon_0] = \frac{[q_1 q_2]}{[F] \cdot [r^2]} = \frac{I^2T^2}{(MLT^{-2}) \cdot (L^2)} \]

### Step 4: Simplifying the Expression
Now we simplify the expression:

1. The denominator becomes:
\[ [F] \cdot [r^2] = MLT^{-2} \cdot L^2 = ML^3T^{-2} \]

2. Therefore, we have:
\[ [\epsilon_0] = \frac{I^2T^2}{ML^3T^{-2}} = \frac{I^2T^2}{M L^3} \cdot T^2 \]

3. Rearranging gives:
\[ [\epsilon_0] = \frac{I^2T^4}{ML^3} \]

### Final Result
Thus, the dimensional formula for permittivity of free space \((\epsilon_0)\) is:

\[ [\epsilon_0] = M^{-1}L^{-3}T^{4}I^{2} \]