Question

A point charge of -1.0

Answer 1

according to coulomb's law,

electric force F = kq₁q₂/r²

where, k = 9*10⁹ N.m²/C²

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from figure, the right angled triangle OAB,

         OA = d = ?[(1)^2 + (0.5)²] = ?1.25 = 1.118

from figure, OD = r

from figure, ?OAB and ?OCD are similar triangles.

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charge q₁ = 16*10^-6 C at (1 m, 0.5 m)

charge q₂ = -1*10^-6 C at (0 m , 0 m)

at equilibrium

the force due to charge q1 exerted on the electron is equals to the

the force due to charge q2 exerted on the electron.

                      F'_1e = F'_2e ............. (1)
             k(q₁) (e)/(AD)² = k(q₂) (e)/(OD)²
             (q₁) /(r+1.118)² = (q₂) /(r)²

             (q₁/q₂) = [ r + 1.118/r] ²  
    [ r + 1.118/r] =  (q₁/q₂)^1/2              .............. (2)

substitute the given data in eq (2), we get

          [ r + 1.118/r] = 4

4*r - r = 1.118

        r = 0.373 m

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from similar triangles ?OAB and ?OCD,

CD/AB = OD/OA

ye/0.5 = 0.373/1.118

y_e = 0.167 m

from similar triangles ?OAB and ?OCD,

OC/OB = OD/OA

xe/1 = 0.373/1.118

x_e = 0.334 m

therefore, position of the electron is (-0.334 m , - 0.167 m)