Zack
Lv 7
Zack asked in Science & MathematicsPhysics · 3 months ago

# 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!

Relevance
• 3 months ago

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.

2.

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

• oldschool
Lv 7
3 months agoReport

d = ½at² = ½(200q/m)t² = 9.8m proton
9.9m electron