Two particles are fixed on an x axis. particle 1 of charge 44.9 μc is cated at х- 24.5 cm; particle 2 of charge q is located at x 7.53 cm. particle 3 o charge magnitude 38.9 uc is released from rest on the y axis at y 24.5 cm, what is the value of q if the initial acceleration of particle 3 is in the positive direction of (a) the x axis and (b) the y axis?

The force that is being exerted on particle 3 by particle 1 is equal to:

As both particles has positive charges, the particles will repel each other and the resultant force will have the direction of the vector . Therefore, will have x and y components equal to:

In order to calculate Force between particles 2 and 3, we first assume Q2 to be possitive (if it's negative the result will have a negative value, so this doesn't matter). We follow the same line of reasoning we used to calculate F13, just that Q2 will be unknown.

a) For incise a, F13y + F23y has to be equal to 0:

b) For incise b, F13x + F23x has to be equal to 0:

568.98

explanation:

its math

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