Two Points charges q1 and q2 are held 4.00 cm apart. An electron released at a p

Question

Two Points charges q1 and q2 are held 4.00 cm

apart. An electron released at a point that is equidistant from both charges undergoes an intial acceleration of 8.10 X 10^18 m/s^2 directly upward in the figure parallel to the line connecting q1 and q2

Find the magnitude and sign of q1 and q2.

Two Points charges q1 and q2 are held 4.00 cm apart. An electron released at a point that is equidistant from both charges undergoes an intial acceleration of 8.10 times 10^18 m/s^2 directly upward in the figure parallel to the line connecting q1 and q2

Explanation / Answer

1stly q1 is +ve and q2 is -ve and they are of equal magnitude. Such an assumption can only cancel the horizontal force and we are left with only vertical force in the direction where acceleration is shown.



Net force acting on the electron = F1 cos (x) + F2 cos (x)
cos (x) = (2 / 3.61) = 0.56

F1 = force on electron due to q1
F2 = force on electron due to q2

F1 = k*(q1 * e) / (0.361 * 0.361) k= 9 * 10^9
e= 1.6 * 10^(-19)
q1 = q
Since magnitude of q1 and q2 are same , so force is also same.

Thus F1 = F2 = F

Hence, net force = 2Fcos (x)

Now [2Fcos (x)] / (mass of the electron ) = acceleration of the electron

[2 * 9 * (10^9) * 1.6 * 10^(-19) * Q * 0.56] / [3.61 * 3.61 * 10^(-4) * 9.1 * 10^(-31)] = 8.1 * 10^(18)

Q = 5.956*10^-6

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