The electric field in a certain region is given by `vec(E) = ((K)/(x^(3)))vec(i)`. The dimensions of K are -

Options

A. `MLT^(-3)A^(-1)`
B. `ML^(-2)T^(-3)A^(-1)`
C. `ML^(4)T^(-3)A^(-1)`
D. dimensionless

Answer

To find the dimensions of the constant \( K \) in the electric field equation \( \vec{E} = \frac{K}{x^3} \vec{i} \), we will follow these steps:

### Step 1: Understand the Electric Field
The electric field \( \vec{E} \) is defined as the force per unit charge. The formula can be expressed as:
\[ \vec{E} = \frac{\vec{F}}{q} \]
where \( \vec{F} \) is the force and \( q \) is the charge.

### Step 2: Determine the Dimensions of Force and Charge
The dimension of force \( \vec{F} \) is given by:
\[ \text{Force} = \text{mass} \times \text{acceleration} = M \cdot L \cdot T^{-2} \]
The dimension of charge \( q \) can be expressed in terms of current \( I \) and time \( T \):
\[ q = I \cdot T \]
Thus, the dimension of charge \( q \) is:
\[ \text{Charge} = A \cdot T \]

### Step 3: Find the Dimensions of Electric Field
Substituting the dimensions of force and charge into the electric field equation gives us:
\[ \text{Dimension of } \vec{E} = \frac{\text{Dimension of Force}}{\text{Dimension of Charge}} = \frac{M \cdot L \cdot T^{-2}}{A \cdot T} = M \cdot L \cdot T^{-3} \cdot A^{-1} \]

### Step 4: Relate Electric Field to K and x
From the given equation \( \vec{E} = \frac{K}{x^3} \), we can express \( K \):
\[ K = \vec{E} \cdot x^3 \]

### Step 5: Find the Dimensions of K
Since \( x \) is a length, its dimension is \( L \). Therefore, the dimension of \( x^3 \) is:
\[ \text{Dimension of } x^3 = L^3 \]
Now substituting the dimensions of \( \vec{E} \) and \( x^3 \) into the equation for \( K \):
\[ \text{Dimension of } K = \text{Dimension of } \vec{E} \cdot \text{Dimension of } x^3 = (M \cdot L \cdot T^{-3} \cdot A^{-1}) \cdot L^3 = M \cdot L^4 \cdot T^{-3} \cdot A^{-1} \]

### Conclusion
Thus, the dimensions of \( K \) are:
\[ \text{Dimension of } K = M \cdot L^4 \cdot T^{-3} \cdot A^{-1} \]