The law goverening electrostatics is coulomb's law.In priciple coulomb's law can be used to compute electric field due to any charge configuation .In pracitce however it is a formidable task to compute electric field due to uniform charge distributions .For such cases Gauss proposed a theorem which states that
`ointbar(E).bar(ds)=(q)/(epsilon_(0))`
where `ds` is an element of area directed across the outward normal for the surface at every point .The integral is called electric flux.Any convenient surface that we choose to evalute the surface integral is called the Gaussian surface.
The `SI` unit of electric flux is

Options

A. weber
B. `"newton coulomb"^(-1)`
C. Volt `xx` metre
D. `"joule coulomb"^(-1)`

Answer

To find the SI unit of electric flux, we start from Gauss's law, which states:

\[ \oint \mathbf{E} \cdot d\mathbf{s} = \frac{q}{\epsilon_0} \]

where:
- \(\mathbf{E}\) is the electric field,
- \(d\mathbf{s}\) is an area element directed outward,
- \(q\) is the charge enclosed,
- \(\epsilon_0\) is the permittivity of free space.

### Step 1: Identify the units of electric field (\(E\))
The electric field \(E\) is defined as the force (\(F\)) per unit charge (\(q\)). The unit of force in SI is Newton (N), and the unit of charge is Coulomb (C). Therefore, the unit of electric field is:

\[ [E] = \frac{N}{C} \]

### Step 2: Identify the units of area (\(d\mathbf{s}\))
The area element \(d\mathbf{s}\) is measured in square meters (m²). Thus, the unit of area is:

\[ [d\mathbf{s}] = m^2 \]

### Step 3: Combine the units to find the unit of electric flux
Electric flux (\(\Phi_E\)) is given by the surface integral of the electric field over an area:

\[ \Phi_E = \oint \mathbf{E} \cdot d\mathbf{s} \]

The unit of electric flux can be derived from the product of the units of electric field and area:

\[ [\Phi_E] = [E] \cdot [d\mathbf{s}] = \left(\frac{N}{C}\right) \cdot (m^2) \]

Substituting the units, we get:

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

### Step 4: Convert Newton to Joules
Recall that 1 Newton (N) is equal to 1 Joule per meter (J/m). Therefore, we can express the unit of electric flux as:

\[ [\Phi_E] = \frac{(J/m) \cdot m^2}{C} = \frac{J \cdot m}{C} \]

### Step 5: Final expression for the unit of electric flux
Thus, the SI unit of electric flux can be expressed as:

\[ [\Phi_E] = \frac{J \cdot m}{C} \]

### Conclusion
The SI unit of electric flux is Joule-meter per Coulomb (J·m/C), which can also be expressed in terms of volts:

\[ [\Phi_E] = V \cdot m \]