If 0, E and t represent the free space permittivity, electric fie… — Solution | JEE Physics

MCQEasyJEE 2026SI System & Derived Units

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
$A/m$
B
$A m^2$
C
$A/m^2$
D
$A m$

Answer

Correct answer:C

Step-by-step solution

Standard Method

Given: $\varepsilon_0$ is free space permittivity, $E$ is electric field, and $t$ is time.

Find: The unit of $\frac{\varepsilon_0 E}{t}$ .

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}

Therefore, the correct option is C and the unit is $A/m^2$ .

Unit Cancellation Trick

Given: $\varepsilon_0$ , $E$ , and $t$ .

Find: The unit of $\frac{\varepsilon_0 E}{t}$ .

Use direct unit cancellation:

\varepsilon_0 \to \frac{\text{C}}{\text{V m}}, \qquad E \to \frac{\text{V}}{\text{m}}

Multiplying gives

\frac{\text{C}}{\text{V m}} \times \frac{\text{V}}{\text{m}} = \frac{\text{C}}{\text{m}^2}

Now divide by $t = \text{s}$ :

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

This works because $\text{V}$ cancels immediately, and $\frac{\text{C}}{\text{s}}$ is the definition of ampere. Therefore, the correct option is C.

Common mistakes

Using the unit of electric field incorrectly as $\text{V m}$ or $\text{V}$ . This is wrong because electric field has unit $\frac{\text{V}}\text{m}$ . Always write $[E] = \frac{\text{V}}{\text{m}}$ before multiplying.
Forgetting to divide by time at the final step. This is wrong because the expression is $\frac{\varepsilon_0 E}{t}$ , not just $\varepsilon_0 E$ . First find $[\varepsilon_0 E]$ and then divide by $\text{s}$ .
Not converting $\frac{\text{C}}{\text{s}}$ into ampere. This is wrong because the final unit should be expressed in standard electrical form. Use $1\, \text{A} = \frac{\text{C}}{\text{s}}$ to get $\frac{\text{A}}{\text{m}^2}$ .

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