If the distance between two charges is halved, the electrical force between them increases by a factor 4.
In fact, the magnitude of the electric force between two charges is given by:

where
k is the Coulomb's constant
q1 and q2 are the two charges
r is the separation between the two charges
We see that the magnitude of the force F is inversely proportional to the square of the distance r. Therefore, if the radius is halved:

the magnitude of the force changes as follows:

so, the force increases by a factor 4.
Answer:
true
Explication:
The acceleration of an object depends on the mass of the object, and the amount of force applied
Inductive reactance (Z) = ω L = 2Πf L = (2Π) (12,000) (L)
I = V / Z
4 A = 16v / (24,000Π L)
Multiply each side by (24,000 Π L):
96,000 Π L = 16v
Divide each side by (96,000 Π) :
L = 16 / 96,000Π = 5.305 x 10⁻⁵ Henry
L = 53.05 microHenry
The answer is 0.025J.
W=1/2*k*x^2
W=1/2*20*0.050^2
W=0.025J