Question

Two charges that are separated by one meter exert 1-N forces on each other. If the charges are pushed together so the separation is 25 centimeters, the force on each charge will be______

Answer 1

If the charges are pushed together so the separation is 25 centimeters, the force on each charge will be 16 N.

Step-by-step explanation:

In electrostatics, the force between two charged particles is given by Coulomb's law. The formula to calculate the force is F = k * (q1 * q2) / r^2, where F is the force, k is the electrostatic constant, q1 and q2 are the charges of the particles, and r is the distance between them.

Using this formula, if the distance between the charges is decreased from 1 meter to 25 centimeters, the force on each charge will increase. The force is inversely proportional to the square of the distance, so when the distance is reduced to 0.25 meters (25 centimeters), the force will be 16 times stronger. Therefore, the force on each charge will be 16 N.

Answer 2

Answer: 16N

Step-by-step explanation:

According to coulombs law which states that the force of attraction between two charges is directly proportional to the product of the charges and inversely proportional to the square of the distance between them. Mathematically,

F = kq1q2/r²

If two charges that are separated by one meter exert 1-N forces on each other, the equation will become

1 = kq1q2/1²

kq1q2 = 1 ... (1)

If the charges are pushed together so the separation is 25 centimeters, the force between them becomes,

F = kq1q2/0.25² (25cm converted to meters)

0.0625F = kq1q2 ... (2)

Dividing equation 1 by 2 to determine the force F on each charges, it becomes;

1/0.0625F = kq1q2/kq1q2

1/0.0625F = 1

0.0625F = 1

F = 1/0.0625

F= 16N