Answer: a) the force will be repulsive
b) the ratio of the new force to the old force will be 2
c) O
Explanation:
a) since charge -Q is moved from A to B, this implies that sphere A is negatively charged. The two spheres are now negatively charged and will repel themselves.
b) initial force will be -q(-Q)/d2
Adding extra charge -Q will cause change on B to become -2Q
The new force will be - 2Q(-q)/d2
Dividing new force by old force will give 2
C) if B is neutralized, the net charge becomes 0 and there will be no force on it.
Answer:
The force per unit length (N/m) on the top wire is 16.842 N/m
Explanation:
Given;
distance between the two parallel wire, d = 38 cm = 0.38 m
current in the first wire, I₁ = 4.0 kA
current in the second wire, I₂ = 8.0 kA
Force per unit length, between two parallel wires is given as;

where;
μ₀ is constant = 4π x 10⁻⁷ T.m/A
Substitute the given values in the above equation and calculate the force per unit length

Therefore, the force per unit length (N/m) on the top wire is 16.842 N/m
Answer:
0 m/sec
Explanation:
b/c they were at rest and initial means at rest ,at rest means 0 HOPE THIS HELPS