Question

Using mass $(M)$, length $(L)$, time $(T)$ and current $(A)$ as fundamental quantities, the dimension of permittivity is

• A. $ML ^{-2} T ^{2} A$
• B. $M ^{-1} L ^{-3} T ^{4} A ^{2}$
• C. $MLT ^{-2} A$
• D. $ML^2 T^{-1} A^2$

Answer 1

Answer: (B)

Putting the dimensions for quantities in the expression containing $\varepsilon_{0}$.
From Coulomb's law, two stationary point charges $q_{1}$ and $q_{2}$ attract/repel each other with a force $F$ which is directly proportional to the product of charges and inversely proportional to the square of distance $r$ between them
That is, $F=\frac{1}{4 \pi \varepsilon_{0}} \frac{q_{1} q_{2}}{r^{2}}$
$\Rightarrow \varepsilon_{0}-\frac{1}{4 \pi} \frac{q_{1} q_{2}}{F r^{2}}$
$\therefore $ Dimensions of permittivity
$\varepsilon_{0}=\frac{\text { dimensions of } q^{2}}{\text { dimensions of } F \times \text { dimensions of } r^{2}}$
$\left[\varepsilon_{0}\right]=\frac{\left[ A ^{2} T ^{2}\right]}{\left[ MLT ^{-2}\right]\left[ L ^{2}\right]}=\left[ M ^{-1} L ^{-3} T ^{4} A ^{2}\right]$