Question

Find the dimensions of the expression \( \dfrac{\varepsilon_0 E}{T} \), where \( \varepsilon_0 \), \(E\), and \(T\) are permittivity, electric field, and time respectively.

• A. \( AL \)
• B. \( AL^{-2} \)
• C. \( MA^{-1}L \)
• D. \( MLA^{2} \)

Hint:

Always cancel dimensions step by step and remember:
Electric field = force per unit charge
Permittivity involves inverse force dimensions

Answer 1

The Correct Option is B

Concept: To find the dimensions of a physical expression, we replace each physical quantity with its dimensional formula and simplify. Relevant dimensional formulas:
Permittivity of free space: \[ [\varepsilon_0] = M^{-1}L^{-3}T^{4}A^{2} \]
Electric field: \[ [E] = \frac{F}{q} = ML T^{-3}A^{-1} \]
Time: \[ [T] = T \]
Step 1: Write the given expression in dimensional form. \[ \left[\frac{\varepsilon_0 E}{T}\right] \]
Step 2: Substitute the dimensions. \[ = \frac{(M^{-1}L^{-3}T^{4}A^{2})(MLT^{-3}A^{-1})}{T} \]
Step 3: Simplify the expression. Combining powers: \[ = M^{0}L^{-2}T^{1}A^{1} \] Dividing by \(T\): \[ = M^{0}L^{-2}A^{1} \]
Step 4: Final dimensions. \[ \boxed{AL^{-2}} \]