Two charges q 1 and q 2 are placed at distance r , so that a Coulomb force F act

Question

Two charges q1 and q2 are placed at distance r, so that a Coulomb force F acts on them. If each of the charges is doubled and the distance is halved, how will the Coulomb force change?

It will increase by a factor of 4.

It will stay the same.

It will decrease by a factor of 8.

It will increase by a factor of 8.

It will decrease by a factor of 16.

It will increase by a factor of 2.

It will increase by a factor of 16.

It will decrease by a factor of 2.

It will decrease by a factor of 4.

In an experiment, we place a test charge inside a box, in order to check whether there is an electric field in the space inside the box. The test charge feels a force, therefore, by definition, an electric field exists inside the box. Does that mean that there must be other charges inside the box too?

No, the electric field that we measured was caused by the test charge itself, according to the formula E=kq/r2.

Yes, the fact that the test charge feels a force means that another charge is there to provide the force, according to the formula F=kq1q2/r2.

Not necessarily: a region of space can have an electric field with no charges present in that region.

Yes, electric fields are created by charges, therefore these charges must be in that box too.

Explanation / Answer

F = kq1q2/r^2

so here if q'1 = 2q, q2'= 2q2 an r' = r/2

then New force F' = k (2q1) (2q2)/r^2/4

F' = 16 F

so answer is   It will increase by a factor of 16.

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formual for Ef is E = Kq/r^2

so Yes, the fact that the test charge feels a force means that another charge is there to provide the force, according to the formula F=kq1q2/r2. is the answer

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