Two questions regarding electric field?
1. Point charges q1 = q2 = 4.0 x 10^-6 C are fixed on the x-axis at x = -3.0m and x = 3.0m. What charge q must be placed at the origin so that the electric field vanishes at x=0, y = 3.0m?
2. An electron and a proton, each starting from rest, are accelerated by the same uniform electric field of 200 N/C. Determine the distance and time for each particle to acquire a kinetic energy of 3.2 x 10^-16 J.
You don't have to solve both of them; just the ones you can. Thanks!
- oldschoolLv 73 months agoFavorite Answer
E = kq/r²
The directions of the fields at 0,3: E1 at 45° and E2 at 135°
r² = 3²+3² = 18 for both charges.
Sum the horizontal components:
Ex = (k*4e-6/18) * cos45 + k*4e-6/18 *cos135
= k*4e-6/18 *[0.7071-0.7071] = 0
Sum the vertical components:
Ey = k*4e-6/18 * sin45 + k*4e-6/18 *sin135
= (k*4e-6/18) *[0.7071+0.7071] = 2824.5 N/C at +90°
Therefore: kq/r² = kq/3² = 2824.5 N/C pointing DOWN (270°)
Thus q must be negative. k = 8.988e9
q = -3²*2824.5/k = -2.83µC or about -2.8µC with 2 s.f.
½mv² = 3.2e-16
(6.4e-16)/m = v² = (at)²
a = F/m = E*q/m = 200q/m
v² = (at)² = 6.4e-16/m = (200q/m)² * t²
6.4e-16/m = (200q/m)² * t²
(6.4e-16)m²/(200q)²m = t²
(6.4e-16)m²/(200*1.602e-19)²m = t²
6.233e17 * m = t²
For a proton t = √(6.233e17*6.674e-11) = 32µs
For an electron t = √(6.233e17*9.109e-31) = 0.75µs