Electric field in a certain region is given by `bar(E)=(A/x^(2)hat i+B/y^(3)hat j)`.The SI unit of `A` and `B` are:

Options

A. `Nm^(2)C^(-1),Nm^(3)C^(-1)`
B. `Nm^(3)C,Nm^(2)C`
C. `Nm^(2)C,Nm^(3)C`
D. `Nm^(3)C^(-1),Nm^(2)C^(-1)`

Answer

To determine the SI units of constants \( A \) and \( B \) in the given electric field expression \( \bar{E} = \left( \frac{A}{x^2} \hat{i} + \frac{B}{y^3} \hat{j} \right) \), we need to analyze the units of the electric field and how they relate to \( A \) and \( B \).

### Step-by-Step Solution:

1. **Understand the Electric Field**:
The electric field \( \bar{E} \) is given in the form of two components:
\[ \bar{E} = \frac{A}{x^2} \hat{i} + \frac{B}{y^3} \hat{j} \]
The unit of electric field \( E \) is Newton per Coulomb (N/C).

2. **Analyze the First Component**:
For the \( \hat{i} \) component:
\[ E_x = \frac{A}{x^2} \]
The units of \( E_x \) must match the units of electric field:
\[ [E_x] = \frac{[A]}{[x^2]} \]
Here, \( [x] \) (the unit of distance) is in meters (m). Thus, \( [x^2] = m^2 \).

Therefore, we have:
\[ [E_x] = \frac{[A]}{m^2} \]
Setting this equal to the unit of electric field:
\[ N/C = \frac{[A]}{m^2} \]
Rearranging gives:
\[ [A] = N \cdot m^2/C \]

3. **Analyze the Second Component**:
For the \( \hat{j} \) component:
\[ E_y = \frac{B}{y^3} \]
Similarly, we have:
\[ [E_y] = \frac{[B]}{[y^3]} \]
Here, \( [y^3] = m^3 \).

Thus:
\[ [E_y] = \frac{[B]}{m^3} \]
Setting this equal to the unit of electric field:
\[ N/C = \frac{[B]}{m^3} \]
Rearranging gives:
\[ [B] = N \cdot m^3/C \]

4. **Final Units**:
- The SI unit of \( A \) is \( N \cdot m^2/C \).
- The SI unit of \( B \) is \( N \cdot m^3/C \).

### Conclusion:
The SI units of \( A \) and \( B \) are:
- \( A: \text{Newton meter}^2 \text{ per Coulomb} \) (N·m²/C)
- \( B: \text{Newton meter}^3 \text{ per Coulomb} \) (N·m³/C)