Two point charges, A and B, are separated by 6 cm. The attractive force between them is 20 N. Find the force between them when they are separated instead by 12 cm. The charges are brought back to their original 6 cm separation, but the magnitude of charge A is doubled. Find the force between them. What would the force become if charge B is doubled as well?

Respuesta :

The force between them when they are separated instead by 12 cm is 5 N.

The force between them when charge A is doubled is 40 N.

The force becomes 80 N  if charge B is doubled as well.

Force between two charges

The force between two charge is determined from Coulomb's law of electrostatic force.

f = kq²/r²

when the distance, r is double the force reduces by 4;

f = kq²/(2r)²

f =  kq²/4r²

New attractive force when the distance is 12cm, = 20 N/4 = 5 N

Magnitude of force when one of the charge is double

f =  k(2q₁)(q₂)/r²

f =  2k(q₁q₂)/r²

New force = 2 x 20 N = 40 N

When charge B is doubled as well

f =  k(2q₁)(2q₂)/r²

f =  4k(q₁q₂)/r²

New force = 4 x 20 N = 80 N

Thus, the force between them when they are separated instead by 12 cm is 5 N.

The force between them when charge A is doubled is 40 N.

The force becomes 80 N  if charge B is doubled as well.

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