Answer:
Electric field acting on the electron is 127500 N/C.
Explanation:
It is given that,
Mass of an electron,
Charge on electron,
Initial speed of electron, u = 0
Final speed of electron,
Distance covered, s = 2 cm = 0.02 m
We need to find the electric field required. Firstly, we will find the acceleration of the electron from third equation of motion as :
According to Newton's law, force acting on the electron is given by :
F = ma
Electric force is given by :
F = q E, E = electric field
E = 127500 N/C
So, the electric field is 127500 N/C. Hence, this is the required solution.
Answer:
B. Mechanical energy= 50J+30J=80J
Answer:
The value of third charge is 0.8μC.
Explanation:
Given that.
Magnitude of net force=4.444 N
According to figure,
Suppose, First charge = 2.4 μC
Second charge = 6.2 μC
Distance r₁ = 9.8 cm
Distance r₂ = 2.1 cm
We need to calculate the value of r
Using Pythagorean theorem
Put the value into the formula
We need to calculate the force
Using formula of force
Force F₁₂,
Force F₂₃,
We need to calculate the value of third charge
Hence, The value of third charge is 0.8μC.
You may look at what group they are in
Group
1A=Group 1
2A = Group 2
3A = Group 13
4A= Group 14
5A=Group 15
6A=Group 16
7A=Group 17
The #A tells you how many valence electrons there are by the # before A. Such as Chlorine, which is in 7A, so therefore has 7 valence electrons.