Answer:
Explanation:
Let A and B be two points located in a uniform electric field, A being a distance d from B in the direction of the field. The work that an external force must do to bring a unit positive charge q from the reference point to the point considered against the electric force at constant speed, mathematically is expressed by:
Therefore, isolating 
and replacing the data provided:
<h3>
Answer: The acceleration doubles</h3>
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Explanation:
Consider a mass of 10 kg, so m = 10
Let's say we apply a net force of 20 newtons, so F = 20
The acceleration 'a' is...
F = ma
20 = 10a
20/10 = a
2 = a
a = 2
The acceleration is 2 m/s^2. Every second, the velocity increases by 10 m/s.
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Now let's double the net force on the object
F = 20 goes to F = 40
m = 10 stays the same
F = ma
40 = 10a
10a = 40
a = 40/10
a = 4
The acceleration has also doubled since earlier it was a = 2, but now it's a = 4.
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In summary, if you double the net force applied to the object, then the acceleration doubles as well.
3.60 A = 3.60 coulombs of charge per second
(3.60 Coul/sec) x (15.3 sec) = 55.08 coulombs of charge
1 coulomb of charge is carried by 6.25 x 10^18 electrons
Number of electrons =
(55.08 Coul) x (6.25 x 10^18 e/coul) = <em>3.4425 x 10^20 electrons</em>
