Question

Force f acts between two charges, q1 and q2, separated by a distance d. if q1 is increased to twice its original value and the distance between the charges is also doubled, what is the new force acting between the charges in terms of f? one quarterf one halff f 2f

Answer 1

The new force acting between the charges in terms of f given the data from the question is half f

Coulomb's law equation

F = Kq₁q₂ / r²

Where

• F is the force of attraction
• K is the electrical constant
• q₁ and q₂ are two point charges
• r is the distance apart

How to determine the initial force

• Charge 1 = q₁
• Charge 2 = q₂
• Electric constant (K) = 9×10⁹ Nm²/C²
• Distance apart (r) = d

• Initial force (F₁) = f =?

F = Kq₁q₂ / r²

F₁ = f = Kq₁q₂ / d²

How to determine the new force

• Charge 1 = 2q₁
• Charge 2 = q₂
• Electric constant (K) = 9×10⁹ Nm²/C²
• Distance apart (r) = 2d

• New force (F₂) =?

F = Kq₁q₂ / r²

F₂ = K2q₁q₂ / (2d²

F₂ = 2Kq₁q₂ / 4d²

F₂ = Kq₁q₂ / 2d²

F₂ = ½(Kq₁q₂ / d²)

But

f = Kq₁q₂ / d²

Therefore

F₂ = ½f = half f

Learn more about Coulomb's law:

https://brainly.com/question/506926

#SPJ4