Q 1.19: A point charge of 2.0\mu Cis at the centre of a cubic Gaussian surface 9.0 cm on edge. What is the net electric flux through the surface?

Answers (1)

Given,

q = net charge inside the cube = 2.0\mu C

Using Gauss’s law, we know that the flux of electric field through any closed surface S is 1/\epsilon _{0}  times the total charge enclosed by S.

i.e.   \phi = q/\epsilon _{0}

where, q = net charge enclosed  and  \epsilon _{0} = permittivity of free space (constant)

\therefore \phi = 2.0\times10^{-6}/8.85\times10^{-12}\ Nm^2C^{-1} = \boldsymbol{2.2\times10^5\ Nm^2C^{-1}}

(Note: Using Gauss's formula, we see that the electric flux through the cube is independent of the position of the charge and dimension of the cube! )

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