Question

If \(\varepsilon_0\), \(E\) and \(t\) represent the free space permittivity, electric field and time respectively, then the unit of \[ \frac{\varepsilon_0 E}{t} \] will be

• A. \( \text{A/m} \)
• B. \( \text{A m}^2 \)
• C. \( \text{A/m}^2 \)
• D. \( \text{A m} \)

Hint:

Always reduce electrical units to Coulomb, second, and meter to simplify dimensional analysis.

Answer 1

The Correct Option is C

To determine the unit of the expression \(\frac{\varepsilon_0 E}{t}\), we need to analyze the dimensional formula of each component in the expression and derive the resultant unit. 

1. Let's consider the dimensional formulae of each component:
   • The permittivity of free space (\(\varepsilon_0\)) has the unit of \(\text{F/m}\ (Farad per meter)\).
   • The electric field (\(E\)) has the unit of \(\text{N/C} = \text{V/m} (Newton per Coulomb or Volt per mete\).
   • Time (\(t\)) is measured in seconds \(s\).
2. Now, substitute these units in the expression \(\frac{\varepsilon_0 E}{t}\):
   • \(\varepsilon_0 E\) has units:

\[\varepsilon_0 \times E = \left(\frac{\text{F}}{\text{m}}\right) \times \left(\frac{\text{V}}{\text{m}}\right) = \frac{\text{F} \cdot \text{V}}{\text{m}^2}\]

• The unit of charge (Coulomb, C) can be related to Farad and Volt as follows: 1 F = 1 C/V, so

\[\text{F} \cdot \text{V} = \frac{\text{C}}{\text{V}} \times \text{V} = \text{C}\]

1. Thus, the unit of \(\varepsilon_0 E\) simplifies to Coulomb per square meter:

\[\varepsilon_0 E = \frac{\text{C}}{\text{m}^2}\]

1. Now, divide by time \(t\):
   • The unit of \(\frac{\varepsilon_0 E}{t}\) becomes:

\[\frac{\text{C}}{\text{m}^2 \cdot \text{s}} = \frac{\text{A}}{\text{m}^2}\]

• This is because \(\text{C/s}\) is equivalent to Ampere (\(\text{A}\)), the unit of current.

Therefore, the unit of \(\frac{\varepsilon_0 E}{t}\) is A/m², which corresponds to the rate of current density.