Question

Two charged particles repel each other with a force F. If the charge of one particle is doubled and the distance between them is doubled, the force will be?A) F.B) 2 F.C) F/2.D) F/4.

Answer 1

C) F/2

Step-by-step explanation

By the Coulomb's Law we know that the force between two charged particles is:

`F=k_e\cdot(q_1\cdot q_2)/(r^2)`

Where r is the distance between the particles, q1 and q2 are the charges and ke is Coulomb's constant.

Now, it is given that the distance between the particles is doubled. Let's call this new distance d:

`d=2r`

And one of the particles now has double as charge as before. Let's say that this is the case of particle 1:

`q_{1\text{new}}=2q_1`

The force in this case is:

`F_{\text{new}}=k_e\cdot\frac{q_{1\text{new}}\cdot q_2}{d^2}`

If we replace with the second and third equations:

`F_{\text{new}}=k_e\cdot(2q_1\cdot q_2)/((2r)^2)=k_e\cdot(2q_1\cdot q_2)/(4r^2)=(1)/(2)k_e\cdot(q_1\cdot q_2)/(r^2)`

Note that what's on the right is the force we initially had for the particles:

`F_{\text{new}}=(1)/(2)F`