A cube with sides of area 32 cm2 contains a 35.9 nanoCoulomb charge. Find the flux of the electric field through the surface of
the cube in unis of Nm2/C.
1 answer:
Answer:
The electric flux through the surface is equal to 3.878 x 10³ Nm²/C
The field distance r is equal to half the length of each side of the cube.
From the area the length of each size was calculated and the field distance and charge were used in calculating the magnitude of the electric field vector which was found to be 202 x 10³ N/C
The total flux area available to this electric field is 6x32cm²
Explanation:
The full solution can be found in the attachment below.
Thank you for reading this post and I hope it is helpful to you.
You might be interested in
Answer:
25 N
Explanation:
Work is a product of force and perpendicular distance moved.
W=Fd where F is force exerted and d is perpendicular distance.
However, for this case, the distance is inclined hence resolving it to perpendicular so that it be along x-axis we have distance as
Therefore,
Making F the subject of the formula then
where
is the angle of inclination. Substituting 190 J for W then 18 degrees for
and 8 m for d then