Explanation:
the answers are the first 3.
Answer:
Explanation:
- given S = distance from the first = 3.20cm = 0.032m, t = 1.30×10−8 s
- acceleration = 0.032 X 2 /(1.30×10−8)^2
a = 3.79 x 10^14m/s^2
E = ma /q = 9.11 x 10^-31 x 3.79 x 10^14 / 1.6 x 10^-19
E = magnitude of this electric field. = 2156.3N/C
b) Find the speed of the electron when it strikes the second plate ; V^2 = 2as
= 2 X 3.79 x 10^14 X 0.032
= 4.92 X 10^6m/s
Speed v = distance travelled / time taken
v = d / t
v = 540 / 60h
v = 9 km /h
Answer:
F = 2.49 x 10⁻⁹ N
Explanation:
The electrostatic force between two charged bodies is given by Colomb's Law:

where,
F = Electrostatic Force = ?
k = colomb's constant = 9 x 10⁹ N.m²/C²
q₁ = charge on proton = 1.6 x 10⁻¹⁹ C
q₂ = second charge = 1.4 C
r = distace between charges = 0.9 m
Therefore,

<u>F = 2.49 x 10⁻⁹ N</u>