Solveeit Logo
Login

Question

Physics Question on Electric charges and fields

A charge Q µc is placed at the centre of cube, the flux coming out from any surfaces will be : -

A

Q6ϵ0×10−6\frac{Q}{6\epsilon_0}\times 10^{-6}

B

Q6ϵ0×10−3\frac{Q}{6\epsilon_0}\times 10^{-3}

C

Q2ϵ0\frac{Q}{2\epsilon_0}

D

Q8ϵ0\frac{Q}{8\epsilon_0}

Answer

Q6ϵ0×10−6\frac{Q}{6\epsilon_0}\times 10^{-6}

Explanation

Solution

The correct option is (A): Q6ϵ0×10−6\frac{Q}{6\epsilon_0}\times 10^{-6}
flux coming out =Qenclosedϵ0\frac{Q_{enclosed}}{\epsilon_0}.
E.ds=Q×10−6ϵ0\frac{Q\times 10^{-6}}{\epsilon_0}(guass law)
=flux by one of the side = Q×10−66ϵ0\frac{Q\times 10^{-6}}{6\epsilon_0}
= Q×10−66ϵ0\frac{Q\times 10^{-6}}{6\epsilon_0}