In Fig. 1, three particles are fixed in place and have charges q 1 = q 2 = +e an
ID: 2108047 • Letter: I
Question
In Fig. 1, three particles are fixed in place and have charges
q1 = q2 = +e and q3 = +2e . Distance a = 6. 00 μ m. What are the (a)
magnitude and (b) direction (relative to the positive x axis) of the net
electric field at point P due to the particles. If a fourth particle of mass
m = 2. 7x
~ 10−30 kg and charge Q = − 3e is placed at P , determine the (c)
magnitude and (d) direction of the net electrostatic force on the charge
Q . (e) What is the magnitude of the initial acceleration of the fourth
particle?
Explanation / Answer
Ep due to q1 and q2 will be equal and opp hence will get cancelled
so Ep = E due to q3 = Kq3/r^2 where r = a(root3)/2 and q3 = 2e and direction will be along the line joining q3 and P
b) F = K q3 x q4 / r^2 where r remains the same abd q3 = 3e and force will be along the line joining q3 and P. the sign of q4 is not clear in the que. if it is negative, then F will be towards q3 and if it is positive, F will be away from q3.
e) once you find out the force, then initial acc = F/m