Answer:
The approximate magnitude of the electric field at the surface of the cube is 1.8 N/C.
Explanation:
The electric field, E, at the surface of the cube can be determined by:
E =[tex]\frac{kq}{r^{2} }[/tex]
Where k is a constant (8.99 x 109 [tex]Nm^{2} C^{-2}[/tex]), q is the charge and r the distance from the charge.
But the volume charge density, ρ, is given as:
ρ = [tex]\frac{q}{V}[/tex]
where: q is the electric charge and V the volume.
From the question, sides of cube = 0.10 m and uniform volume charge density, ρ = 2.0 × [tex]10^{-9}[/tex][tex]Cm^{-3}[/tex].
Thus:
volume of cube = length ×length × length
= 0.10 ×0.10 ×0.10
= 1.0 × [tex]10^{-3}[/tex] [tex]m^{3}[/tex]
So that:
q = ρ × V
= 2.0 × [tex]10^{-9}[/tex] × 1.0 × [tex]10^{-3}[/tex]
= 2.0 × [tex]10^{-12}[/tex]
q = 2.0 × [tex]10^{-12}[/tex] C
The electric field, E, at the surface of the cube can be determined by:
E =[tex]\frac{kq}{r^{2} }[/tex]
k = 8.99 x 109 [tex]Nm^{2} C^{-2}[/tex]), q = 2.0 × [tex]10^{-12}[/tex] C, r = 0.1 m, so that:
E = [tex]\frac{8.99*10^{9} * 2*10^{-12} }{0.1^{2} }[/tex]
= 1.798 N/C
The approximate magnitude of the electric field at the surface of the cube is 1.8 N/C.
