Answer: 14.52*10^6 m/s
Explanation: In order to explain this problem we have to consider the energy conservation for the electron within the coaxial cylidrical wire.
the change in potential energy for the electron; e*ΔV is equal to energy kinetic gained for the electron so:
e*ΔV=1/2*m*v^2 v^=(2*e*ΔV/m)^1/2= (2*1.6*10^-19*600/9.1*10^-31)^1/2=14.52 *10^6 m/s
Answer:
Statement 1) False
Statement 2) False
Statement 3) True
Explanation:
The uncertainty principle states that " in a physical system certain quantities cannot be measured with random precision no matter whatever the least count of the instrument is" or we can say while measuring simultaneously the position and momentum of a particle the error involved is

Thus if we measure x component of momentum of a particle with 100% precision we cannot measure it's position 100% accurately as the error will be always there.
Statement 1 is false since measurement of x and y positions has no relation to uncertainty.
Statement 2 is false as both the momentum components can be measured with 100% precision.
Statement 3 is true as as demanded by uncertainty principle since they are along same co-ordinates.
Answer:
q = 4.5 nC
Explanation:
given,
electric field of small charged object, E = 180000 N/C
distance between them, r = 1.5 cm = 0.015 m
using equation of electric field

k = 9 x 10⁹ N.m²/C²
q is the charge of the object

now,

q = 4.5 x 10⁻⁹ C
q = 4.5 nC
the charge on the object is equal to 4.5 nC
Answer:
E/4
Explanation:
The formula for electric field of a very large (essentially infinitely large) plane of charge is given by:
E = σ/(2ε₀)
Where;
E is the electric field
σ is the surface charge density
ε₀ is the electric constant.
Formula to calculate σ is;
σ = Q/A
Where;
Q is the total charge of the sheet
A is the sheet's area.
We are told the elastic sheet is a square with a side length as d, thus ;
A = d²
So;
σ = Q/d²
Putting Q/d² for σ in the electric field equation to obtain;
E = Q/(2ε₀d²)
Now, we can see that E is inversely proportional to the square of d i.e.
E ∝ 1/d²
The electric field at P has some magnitude E. We now double the side length of the sheet to 2L while keeping the same amount of charge Q distributed over the sheet.
From the relationship of E with d, the magnitude of electric field at P will now have a quarter of its original magnitude which is;
E_new = E/4