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The dimension of permittivity $(\varepsilon _0)$ are ______. Take Q as the dimension of charge.A) $[M^{-1} L^{-2} T^{-2} Q^{-2}]$ B) $[M^{-1} L^{-3} T^{2} Q^{2}]$ C) $[M^{-1} L^{2} T^{-3} Q^{-1}]$ D) $[M^{-1} L^{3} T^{-2} Q^{-2}]$

Last updated date: 22nd Jun 2024
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Hint: The dimensions of a physical quantity are the power raised by the basic units to obtain a unit of those quantities. For derived quantity represented by ${Q = }{{{M}}^{{a}}}{{{L}}^{{b}}}{{{T}}^{{c}}}$, then ${{{M}}^{{a}}}{{{L}}^{{b}}}{{{T}}^{{c}}}$ is called dimensional formula.

Complete solution:
${F = }\dfrac{{K}{{{Q}}^{{2}}}}{{{{r}}^{{2}}}}$
and dimension of $F = [M L T^{-2}]$
$\Rightarrow {ML}{{{T}}^{-2}}{ = }\dfrac{{K}{{{Q}}^{{2}}}}{{{{r}}^{{2}}}}$
$\Rightarrow K = [M L^3 T^{-2} Q^{-2}]$
Substituting value of K
$\Rightarrow \dfrac{1}{4 \pi \varepsilon _0}$

Hence, answer is option (B), $[M^{-1} L^{-3} T^2 Q^2]$.

Note: Vacuum refers to the minimum possible value of permittivity allowed. This is commonly known as free space or permitted electronic constant. Denoted by $\varepsilon _0$ and has the value $8.85 \times {10}^{-12} Farad/meter$. Disputes against the formation of electric field lines are also evident in the indictment. The permittivity of a dielectric is represented by the ratio of its absolute permittivity to the electronic constant and is usually given relative to the erase.