The electric field is given by `vecE=(A)/(x^(3))hati+By hatj+Cz^(2)hatk`. The SI units of A, B and C are respectively:

Options

A. `(N-m^(3))/(C ), V//m^(2), N//m^(2)-C`
B. `V-m^(2), V//m, N//m^(2)-C`
C. `V//m^(2), V//m, N-C//m^(2)`
D. `V//m, N-m^(3)//C, N-C//m`

Answer

To find the SI units of A, B, and C in the electric field expression given by \(\vec{E} = \frac{A}{x^3} \hat{i} + B y \hat{j} + C z^2 \hat{k}\), we will analyze each term of the electric field.

### Step 1: Understand the electric field
The electric field \(\vec{E}\) is a vector quantity, and its SI unit is Newton per Coulomb (N/C) or equivalently, Volt per meter (V/m).

### Step 2: Analyze the first term \(\frac{A}{x^3}\)
In the first term, we have \(\frac{A}{x^3}\). For this term to have the same units as the electric field \(\vec{E}\), we can set up the equation:

\[ \frac{A}{x^3} \text{ (units)} = \text{Electric Field (N/C)} \]

Where \(x\) is a distance measured in meters (m). Thus, the units of \(x^3\) are:

\[ [x^3] = m^3 \]

Rearranging the equation gives us:

\[ A = \text{Electric Field} \times x^3 = \left(\frac{N}{C}\right) \times m^3 \]

Thus, the SI unit of \(A\) is:

\[ [A] = \frac{N \cdot m^3}{C} \]

### Step 3: Analyze the second term \(B y\)
In the second term, we have \(B y\). Again, for this term to have the same units as the electric field, we can set up the equation:

\[ B y \text{ (units)} = \text{Electric Field (N/C)} \]

Where \(y\) is also a distance measured in meters (m). Thus, the units of \(y\) are:

\[ [y] = m \]

Rearranging gives us:

\[ B = \frac{\text{Electric Field}}{y} = \frac{N/C}{m} \]

Thus, the SI unit of \(B\) is:

\[ [B] = \frac{N}{C \cdot m} \]

### Step 4: Analyze the third term \(C z^2\)
In the third term, we have \(C z^2\). Again, for this term to match the electric field, we set up the equation:

\[ C z^2 \text{ (units)} = \text{Electric Field (N/C)} \]

Where \(z\) is a distance measured in meters (m). Thus, the units of \(z^2\) are:

\[ [z^2] = m^2 \]

Rearranging gives us:

\[ C = \frac{\text{Electric Field}}{z^2} = \frac{N/C}{m^2} \]

Thus, the SI unit of \(C\) is:

\[ [C] = \frac{N}{C \cdot m^2} \]

### Summary of SI Units
- The SI unit of \(A\) is \(\frac{N \cdot m^3}{C}\).
- The SI unit of \(B\) is \(\frac{N}{C \cdot m}\).
- The SI unit of \(C\) is \(\frac{N}{C \cdot m^2}\).