Explanation:
It is given that,
Mass of an electron, 
Initial speed of the electron, 
Final speed of the electron, 
Distance, d = 5 cm = 0.05 m
(a) The acceleration of the electron is calculated using the third equation of motion as :



Force exerted on the electron is given by :



(b) Let W is the weight of the electron. It can be calculated as :



Comparison,


Hence, this is the required solution.
Answer:
-4.71 m/s
Explanation:
Given:
y₀ = 1.13 m
y = 0 m
v₀ = 0 m/s
a = -9.8 m/s²
Find: v
v² = v₀² + 2a (y − y₀)
v² = (0 m/s)² + 2(-9.8 m/s²) (0 m − 1.13 m)
v = -4.71 m/s
Answer:
y = 10.2 m
Explanation:
It is given that,
Charge, 
It is placed at a distance of 9 cm at x axis
Charge, 
It is placed at a distance of 16 cm at x axis
We need to find the point on the y-axis where the electric potential zero. The net potential on y-axis is equal to 0. So,

Here,

So,

Squaring both sides,

So, at a distance of 10.2 m on the y axis the electric potential equals 0.