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

Step 1: Write the units of each quantity.
Free space permittivity: \[ [\varepsilon_0] = \frac{\text{C}}{\text{V m}} \] Electric field: \[ [E] = \frac{\text{V}}{\text{m}} \] Time: \[ [t] = \text{s} \]
Step 2: Find the unit of \(\varepsilon_0 E\).
\[ [\varepsilon_0 E] = \frac{\text{C}}{\text{V m}} \times \frac{\text{V}}{\text{m}} = \frac{\text{C}}{\text{m}^2} \]
Step 3: Divide by time.
\[ \left[\frac{\varepsilon_0 E}{t}\right] = \frac{\text{C}}{\text{m}^2 \text{s}} \] Since \[ 1\ \text{A} = \frac{\text{C}}{\text{s}}, \] we get \[ \left[\frac{\varepsilon_0 E}{t}\right] = \frac{\text{A}}{\text{m}^2}. \]
Final Answer: \[ \boxed{\text{A/m}^2} \]