The initial force between the two charges is given by:

where k is the Coulomb's constant, q1 and q2 the two charges, d their separation. Let's analyze now the other situations:
1. F
In this case, q1 is halved, q2 is doubled, but the distance between the charges remains d.
So, we have:

So, the new force is:

So the force has not changed.
2. F/4
In this case, q1 and q2 are unchanged. The distance between the charges is doubled to 2d.
So, we have:

So, the new force is:

So the force has decreased by a factor 4.
3. 6F
In this case, q1 is doubled and q2 is tripled. The distance between the charges remains d.
So, we have:

So, the new force is:

So the force has increased by a factor 6.
Answer:
1 joule = 0.737 foot-pound
Joule is the unit of work.
1 J = 1 N·m
In SI units
1 J = 1 kg· m/s²
0.737 foot-pound is the amount of work to raise 0.737 pounds one foot or raising one pound to 0.737 ft.
Answer:
the capacitor voltage is V = 20 V
Explanation:
Given,
Capacitance of the capacitor = 2.0 μF
energy stored = 200 W
time (t) =2.0 μs
Capacitor voltage = ?



we know,




V = 20 V
so , the capacitor voltage is V = 20 V
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