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A positive test charge q is released from rest at distance r away from a charge of Q and a distance 2r away from a charge of 2Q. How will the test charge move immediately after being released?

to the left
to the right
stay still
other
Question) 2) Briefly explain your reasoning

User MKiss
by
6.0k points

1 Answer

7 votes

Answer: Option (b) is the correct answer.

Step-by-step explanation:

It is given that a positive test charge q is released from rest at a distance r away from a charge of +Q and a distance 2r which is away from a charge of +2Q.

Then test charge to the right immediately after being released.

Therefore, the net force will be as follows.

F =
(kqQ)/(r^(2)) - kq((2Q))/((2r)^(2))

=
(4KqQ - 2KqQ)/(4r^(2))

=
(KqQ)/(2r^(2))

F =
(KqQ)/(2r^(2)) > 0

Thus, we can conclude that the test charge move to the right immediately after being released.

User Jeremy Hutchinson
by
5.9k points