Two point charges q1 and q2 are held in place 4.50 cm apart. another point charge q = -1.85 mc, of mass 5.50 g, is initially located 3.00 cm from both of these charges and released from rest. you observe that the initial acceleration of q is 314 m> s2 upward, parallel to the line connecting the two point charges. find q1 and q2.

The values of charge 1 and charge 2 are  and   respectively.

Given data:

The distance between the charges is, d = 4.50 cm = 0.045 m.

The magnitude of third charge is, .

The mass of third charge is, m = 5.50 g = 0.00550 kg.

The distance of third charge from other charges is, d' = 3.00 cm = 0.03 m.

The acceleration of third charge is, .

The force on third charge due to other charges is,

Since, the above force is of same in magnitude but opposite to electrostatic force on charge and

Therefore,

The force between the third charge and second charge is,

Thus, the values of charge 1 and charge 2 are  and   respectively.

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Explanation:

Force on the charge Q = -1.85 mC is along the line joining the two charges

so here we can say that net force on it is given by

Now this is the force due to two charges which are in same magnitude but opposite sign

so this force is given as

here we know that

here we know that

now we have

c

explanation:

the maxium height is 34m/s