Understanding the Unit of Electric Flux in SI Base Units

The question asks for the unit of electric flux expressed entirely in terms of the base units of the International System of Units (SI). Electric flux ($\Phi_E$) is a measure of the electric field passing through a given area. It is defined mathematically by the integral of the electric field ($\vec{E}$) dotted with the differential area vector ($d\vec{A}$):

\(\Phi_E = \int \vec{E} \cdot d\vec{A}\)

For a uniform electric field ($\vec{E}$) and a flat area ($\vec{A}$), the electric flux is given by:

\(\Phi_E = E A \cos\theta\)

Where E is the magnitude of the electric field, A is the area, and \(\theta\) is the angle between the electric field vector and the area vector.

Deriving the Electric Flux Unit

To find the unit of electric flux, we need to find the units of electric field and area in SI base units and then multiply them. The SI base units are kilogram (kg), meter (m), second (s), ampere (A), Kelvin (K), mole (mol), and candela (cd). Electric flux involves electric field and area, which relate to mechanical quantities (mass, length, time) and electrical quantities (current).

Unit of Area (A)

Area is a measure of surface extent. The SI unit of length is the meter (m). Area is length multiplied by width, so its unit is \(\text{m} \times \text{m} = \text{m}^2\). This is already in terms of a base unit.

Unit of Electric Field (E)

The electric field is defined as the force per unit charge:

\(E = \frac{F}{q}\)

So, the unit of electric field is the unit of force divided by the unit of charge.

• Unit of Force (F): Force has the SI unit of Newton (N). From Newton's second law (\(F=ma\)), where \(m\) is mass and \(a\) is acceleration, the unit of force can be expressed in base units: \(\text{kg} \times \frac{\text{m}}{\text{s}^2} = \text{kg} \cdot \text{m} \cdot \text{s}^{-2}\).
• Unit of Charge (q): Electric charge has the SI unit of Coulomb (C). Charge is related to electric current (\(I\)) and time (\(t\)) by the definition of current: \(I = \frac{q}{t}\). Rearranging for charge, \(q = I \cdot t\). The SI base unit for current is the ampere (A), and for time is the second (s). Therefore, the unit of charge in base units is \(\text{A} \cdot \text{s}\).

Now, substitute the base units for force and charge into the expression for the unit of electric field:

Unit of \(E = \frac{\text{Unit of Force}}{\text{Unit of Charge}} = \frac{\text{kg} \cdot \text{m} \cdot \text{s}^{-2}}{\text{A} \cdot \text{s}}\)

Simplifying the exponents, we get the unit of electric field in base units:

Unit of \(E = \text{kg} \cdot \text{m} \cdot \text{s}^{-2} \cdot \text{A}^{-1} \cdot \text{s}^{-1} = \text{kg} \cdot \text{m} \cdot \text{s}^{-3} \cdot \text{A}^{-1}\)

Unit of Electric Flux (\(\Phi_E\))

The unit of electric flux is the unit of electric field multiplied by the unit of area:

Unit of \(\Phi_E\) = Unit of \(E \times\) Unit of \(A\)

Unit of \(\Phi_E\) = \((\text{kg} \cdot \text{m} \cdot \text{s}^{-3} \cdot \text{A}^{-1}) \times (\text{m}^2)\)

Combining the terms with the same base (m):

Unit of \(\Phi_E\) = \(\text{kg} \cdot \text{m}^{1+2} \cdot \text{s}^{-3} \cdot \text{A}^{-1}\)

Unit of \(\Phi_E\) = \(\text{kg} \cdot \text{m}^{3} \cdot \text{s}^{-3} \cdot \text{A}^{-1}\)

So, the unit of electric flux in terms of SI base units is \(\text{kgm}^3\text{s}^{-3}\text{A}^{-1}\).

Comparing with Options

Let's compare our derived unit \(\text{kgm}^3\text{s}^{-3}\text{A}^{-1}\) with the given options:

• Option 1: \(\text{kg}^{-1}\text{m}^3\text{s}^{-3}\text{A}^{-1}\) - Incorrect (mass has positive exponent)
• Option 2: \(\text{kgm}^3\text{s}^3\text{A}^{-1}\) - Incorrect (time exponent is positive)
• Option 3: \(\text{kgm}^3\text{s}^{-3}\text{A}\) - Incorrect (current exponent is positive)
• Option 4: \(\text{kgm}^3\text{s}^{-3}\text{A}^{-1}\) - Correct (matches our derivation)

The derived SI base unit for electric flux matches option 4.

Revision Table: Key Physical Quantities and Their SI Base Units

Additional Information: Electric Flux and Gauss's Law

Electric flux is a fundamental concept in electromagnetism. It is closely related to Gauss's Law, which states that the total electric flux through any closed surface (a Gaussian surface) is proportional to the enclosed electric charge. Mathematically, Gauss's Law is given by:

\(\oint \vec{E} \cdot d\vec{A} = \frac{q_{enclosed}}{\epsilon_0}\)

Where \(\oint \vec{E} \cdot d\vec{A}\) is the electric flux through the closed surface, \(q_{enclosed}\) is the total charge enclosed by the surface, and \(\epsilon_0\) is the permittivity of free space. The unit of \(\epsilon_0\) can also be expressed in SI base units. Since \(\epsilon_0\) has units of \(C^2 / (N \cdot m^2)\), its base units are \((A \cdot s)^2 / ((kg \cdot m \cdot s^{-2}) \cdot m^2) = A^2 \cdot s^2 / (kg \cdot m^3 \cdot s^{-2}) = kg^{-1} \cdot m^{-3} \cdot s^4 \cdot A^2\). You can see that the unit of flux (Charge / \(\epsilon_0\)) is indeed \((A \cdot s) / (kg^{-1} \cdot m^{-3} \cdot s^4 \cdot A^2) = kg \cdot m^3 \cdot s^{1-4} \cdot A^{1-2} = kg \cdot m^3 \cdot s^{-3} \cdot A^{-1}\), which confirms our earlier derivation of the electric flux unit.

Understanding how to break down derived physical units into SI base units is a crucial skill in physics, helping to verify equations and understand relationships between different physical quantities.