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# a point charge 2*10^-7C is placed at center of a cube of 0.60 metre side.what is the electric flux through its surfaces

Vikas TU
14149 Points
4 years ago
electric flux = E*dS
Surface are is gven for a face of cube that is=> a^2 => 0.36 m^2

But from Gauss’s law we know that
E*dS = qin/e0
Hence,
flux = 2*10^-7/e0 for the 6 faces included that if for total surface are.
For one surface the net flux would become =>  2*10^-7/6*e0
=> 10^-7/3e0.

Kaur ritika
11 Points
3 years ago
From guass law-electric flux=q/e0So there are total six surfaces and we have to calculate electric flux through six surfacesSo acc to gauss lawElectric flux=q/e0=6q÷6e0=q÷e0=2*10raise to power -7÷0.36=5.55*10raise to power -7.