Physics, asked by aditi130420, 11 months ago

two charges separated by a distance d experiences a force F. if each charge is doubled and distance between them is halved, then what does the force between them become?

Answers

Answered by dhruvsh
1

Answer:

The electric force without the magnetic field, that is the reduced Lorentz Force equation, which is also known as the Coulomb force.

So, Coulomb force is proportional to the product of the magnitude of the two charges and inversely proportional to the square of the distance between the two point charges.

So,

F is proportional to q1q2/r^2

Now, r -> r/2(as the distance is halved)

And,

q1->2q1 and q2-> 2q2 as the magnitude of the two charges are doubled

Therefore,the new force will be

F proportional to 2q1*2q2/(r/2)^2 = 4*4 (q1q2/r^2)

Therefore, the new force after the modifications will be 16 times stronger than the earlier force.

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