A 2-C and a 4-C charge attract each other with a 10-N froce when placed a certain distance apart. With how much force will a 2-C charge and an 8-C charge attract each other when placed the same distance apart?

** I know the answer, but can't find out how to solve it?? **

F=kqq/d^2

so you increase qq by 2x? Has to double force.

To solve this problem, we can use Coulomb's law, which states that the force of attraction or repulsion between two charges is directly proportional to the product of their magnitudes and inversely proportional to the square of the distance between them.

Coulomb's law can be mathematically represented as:

F ∝ (q1 * q2) / r^2

where F is the force of attraction or repulsion, q1 and q2 are the magnitudes of the charges, and r is the distance between them.

In this case, we are given that a 2-C charge and a 4-C charge attract each other with a 10-N force when placed a certain distance apart. Let's call this distance r. So, we can write this as:

10 N ∝ (2 C * 4 C) / r^2

Simplifying the equation further:

10 N ∝ (8 C^2) / r^2

Now, we can find the proportionality constant in this equation. We can do this by rearranging the equation:

10 N = k * (8 C^2) / r^2

where k is the proportionality constant.

To find the force of attraction between a 2-C charge and an 8-C charge when placed the same distance apart, we need to substitute the new charges into the equation. Let's call the force F2:

F2 = k * (2 C * 8 C) / r^2

Now, we need to find the proportionality constant k. To do this, we can rearrange the equation once again:

k = (10 N * r^2) / (8 C^2)

Now, substitute this value of k back into the equation for F2:

F2 = [(10 N * r^2) / (8 C^2)] * (2 C * 8 C) / r^2

Simplifying the equation further:

F2 = (10 N * 2 * 8) / 8

F2 = 20 N

Therefore, the 2-C charge and the 8-C charge will attract each other with a force of 20 N when placed the same distance apart.