Question

Force between two charged is 32.81 if charge one of object is double a distance between them is double what is new force

2nd question: force between two charged is 46.66 if distance between the objects is double what is the new force

Answer 1

Step-by-step explanation:

Eforce = C q1 q2 / r^2 now double q1 and double r

Eforce = C 2 q1 q2 / (2r)^2

EForce = C 2 q1 q2 / (4 r^2)

EForce = 2/4 C q1 q2 / r^2 <=====this shows the NEW force is 1/2 the original

1/2 * 32.81 = 16.41 units

2nd) Similarly this will result in 1/4 the original force

1/4 * 46.66 = 11.67 units

Answer 2

First question:

If the charge of one of the objects is double and the distance between them is also double, then the new force can be found using Coulomb's law:

F = k * (q1 * q2) / r^2

where F is the force between the charges, k is the Coulomb's constant, q1 and q2 are the charges, and r is the distance between them.

Let's assume that the charges on the two objects are q1 and q2, with q1 being double the charge of q2. The distance between them is 2r, where r is the original distance.

Using the formula, we can write:

F = k * (q1 * q2) / r^2

The new force F' can be found by substituting q1 with 2q2 and r with 2r:

F' = k * (2q2 * q2) / (2r)^2

F' = k * 2 * q2^2 / 4r^2

F' = k * q2^2 / 2r^2

The new force is equal to half the original force, or 16.405 N (assuming the original force is in Newtons).

Second question:

If the distance between the objects is doubled, the new force can be found using the inverse-square law of Coulomb's law:

F' = F * (r / 2r)^2

where F is the original force and r is the original distance.

Simplifying this equation, we get:

F' = F / 4

Therefore, the new force is one-fourth of the original force. In this case, the new force would be 46.66 / 4 = 11.665 N (assuming the original force is in Newtons).