# More physics questions.?

1. The charged particle (+q) is in the homogeneous electric field and is affected by the electric force. What happens to the magnitude and direction of the electric force when

- the charge “q” is doubled

- the charge “q” is reversed ( I guess it is from “ + “ to “ - “ )

So, it is that if the charge “+q” is doubled then the magnitude also doubles, but the direction does not change.

So, if the charge “+q” is reversed to "-q" then the magnitude is the same, but the direction is opposite now.?

2. Why the metal can, which is brought near the the positively charged rod, begins to pull it. ?

Relevance
• basically like charges repel

when you bring a metal can near a positively charged rod..

this is what happenes

the rod stays positive ..

the can was neutral meaning the electrons and protons are evenly distributed

as you bring the can near the rod

the electrons from one side of the can are attracted to the other side because the metal is a conductor the electrons can move freely this is what the other person might have meant by inducing a voltage ..

since now the electrons ll move to the side closest the rod the can effectivly becomes negatively charged on the near side to the rod ..

the other side of the can is positively charged but the electrons shield the positive charges from the electric field of the rod

Can

++..--...........Rod

++..--............+

++..--............+

++..--............+

++..--............+

you can see that the effect of this is to make the can negative and attract it to the rod

the positive charges cant move the ones on the right side are masked by the orignial electrons there ..

so i represented the charge on the right as only the free extra electrons that moved ..

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• Anonymous
9 months ago

For the first question, the answers look correct. The equation for electric force F depends on an electric field's strength E and the charge of the particle that the force acts on q.

F = Eq

(If the electric field E is caused by another particle with charge Q, then this would be (E = kQ/r^2)

F = kQq/r^2)

The answer to the second question is probably that the charged rod "induces" a the charges in the metal can to polarize, which sometimes happens in conductors when a charge is near. Basically, this means that the electrons in the metal can flow toward the side of the can that is near the rod.

This is explained in this video:

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