The dimensions of permittivity varepsilon-0are

Question

Toppr Admin · Accepted Answer

Correct option is B- -A-2T-4M-1-L-3-The electrostatic force is given as-F-dfrac-1-4 -pi -varepsilon-0-dfrac-q-1q-2-r-2-Rightarrow -varepsilon-0-dfrac-q-1-q-2-F-r-2-dfrac-A-2T-2-MLT-2-L-2-A-2T-4M-1-L-3-

The dimensions of permittivity $$\varepsilon_0$$are

Correct option is B. $$A^2T^4M^{-1}L^{-3}$$
The electrostatic force is given as:
$$F=\dfrac{1}{4 \pi \varepsilon_0}\dfrac{q_1q_2}{r^2}$$

$$\Rightarrow \varepsilon_0=\dfrac{|q_1||q_2|}{[F][r^2]}$$

$$=\dfrac{[A^2T^2]}{[MLT^{-2}][L^2]}$$

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

The dimensions of permittivity $$\varepsilon_0$$are

Correct option is B. $$A^2T^4M^{-1}L^{-3}$$The electrostatic force is given as:$$F=\dfrac{1}{4 \pi \varepsilon_0}\dfrac{q_1q_2}{r^2}$$$$\Rightarrow \varepsilon_0=\dfrac{|q_1||q_2|}{[F][r^2]}$$$$=\dfrac{[A^2T^2]}{[MLT^{-2}][L^2]}$$$$=[A^2T^4M^{-1}L^{-3}]$$

Correct option is B. $$A^2T^4M^{-1}L^{-3}$$
The electrostatic force is given as:
$$F=\dfrac{1}{4 \pi \varepsilon_0}\dfrac{q_1q_2}{r^2}$$

$$\Rightarrow \varepsilon_0=\dfrac{|q_1||q_2|}{[F][r^2]}$$

$$=\dfrac{[A^2T^2]}{[MLT^{-2}][L^2]}$$

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