Problem 10-31(a) Vector electric field - Part 5 - D
The figure shows the positions of two a particles (charge on each is \(+2e\)) and an electron. What are the magnitude and direction of
(a) the resultant force on the electron?
(Hint: choose the \(+x\) axis along the line from the electron to the right-hand a particle.)
[Ans.(a) \(1.31 \times 10^{-8} N\) at \(13.3^\circ\) to left of line from electron to right-hand a]
Accumulated Solution
\(F_{1x} = F_1, F_{2x} = F_2 \cos 53.1 \\ F_{1y} = 0, F_{2y} = F_2 \sin 53.1\)
Correct:
To evaluate the total \(x\) and \(y\) components we need the magnitudes of the forces \(F_1\) and \(F_2.\) These are calculated from:
(A) \(F = qE\)
(B) \(F = ma\)
(C) \(F = kq_1q_2/r^2\)