The dimensions of permittivity {varepsilon _0 | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) According to Coulomb's Law,the force $F$ between two charges $q_1$ and $q_2$ separated by a distance $r$ is given by: $F = \frac{1}{{4\pi {\vareps

Vedclass · Accepted Answer

${A^2}{T^4}{M^{ - 1}}{L^{ - 3}}$

Vedclass · Answer

(B) According to Coulomb's Law,the force $F$ between two charges $q_1$ and $q_2$ separated by a distance $r$ is given by: $F = \frac{1}{{4\pi {\varepsilon _0}}}\,\frac{{{q_1}{q_2}}}{{{r^2}}}$Rearranging the formula to solve for permittivity ${\varepsilon _0}$:${\varepsilon _0} = \frac{{{q_1}{q_2}}}{{4\pi F{r^2}}}$Substituting the dimensions of each quantity:$[q] = [AT]$$[F] = [MLT^{-2}]$$[r] = [L]$Therefore,the dimensions of ${\varepsilon _0}$ are:$[{\varepsilon _0}] = \frac{[AT][AT]}{[MLT^{-2}][L^2]}$$[{\varepsilon _0}] = \frac{[A^2T^2]}{[ML^3T^{-2}]}$$[{\varepsilon _0}] = [A^2T^4M^{-1}L^{-3}]$Thus,the correct option is $B$.

The dimensions of permittivity ${\varepsilon _0}$ are

Similar Questions

If the dimensions of a physical quantity are given by $[L^{a} M^{b} T^{c}]$,then the physical quantity is:

The dimensional formula for $r.m.s.$ (root mean square) velocity is

The dimensional formula for volume elasticity is:

If $e$ is the electronic charge,$c$ is the speed of light in free space,and $h$ is Planck's constant,the quantity $\frac{1}{4 \pi \varepsilon_{0}} \frac{| e |^{2}}{h c}$ has dimensions of .......

The dimensional formula for impulse is

Vedclass Test Series

Exam Paper Generator

Online Exam Module