Problem 10-31(a) Vector electric field - Part 6 - C

Diagram indicating the positions of two particles.

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

Coordinate diagram B Coordinate System C .

\(F_{1x} = F_1, F_{2x} = F_2 \cos 53.1 \\ F_{1y} = 0, F_{2y} = F_2 \sin 53.1\)

Correct.

That is Coulomb's Law. The magnitude of these forces is:

Solution \(F_1 N\) \(F_2 N\)
(A) \(1.05 \times 10^{-8}\) \(3.76 \times 10^{-9}\)
(B) \(5.22 \times 10^{-9}\) \(1.88 \times 10^{-9}\)