Question

In the figure particle 1 of charge q1 = 0.90 μC and particle 2 of charge q2 = -3.07 μC, are held at separation L = 9.0 cm on an x axis. If particle 3 of unknown charge q3 is to be located such that the net electrostatic force on it from particles 1 and 2 is zero, what must be the (a)x and (b)y coordinates of particle 3?

Answer 1

Let's place q₁ at origin.

Let q₃ be a charge on x-axis. Hence it's y-coordinate will be zero.

Let X be the distance from q₁ to q₃ .

Electrostatic force of attraction between q₁and q₃ is given by:

Fi = .91.43 X2

Similarly, Electrostatic force of attraction between q₂and q₃ is given by:

F = K. 42.43 2-K.(2-x)2

We have,

Fi + F2 = 0

F1 = -2

12(x-7).

Cancelling out K and q₃ :

)

Substituting given values (be sure to include the same units):

0.9 X² -3.07 (9-x)2)

0.9 x2 3.07 (9-x)2

3.07x2 = 0.9 x (9-x)2

3.07x2 = 0.9 x (81 - 18x + x)

3.07x2 = (72.9- 16.2x + 0.9x2)

2.17X2 + 16,2x - 72.9 = 0

Solving for X we get (in cm) :

X = 3.16 cm

That is the co-ordinate of the charge q₃ (3.16 , 0) cm or (0.0316 , 0 ) m

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