Answer:
(a) Charge density σ=6.6375×10²nC/m²
(b) Total charge Q=1.47×10²nC
Explanation:
Given Data
A=47.0 cm =0.47 m
Electric field E=75.0 kN/C
To find
(a) Charge density σ
(b)Total Charge Q
Solution
For (a) charge density σ
From Gauss Law we know that
Φ=Q/ε₀.......eq(i)
Where
Φ is electric flux
Q is charge
ε₀ is permittivity of space
And from the definition of flux
Φ = EA
The flux is electric field passing perpendicularly through the surface
Put the this Φ in equation(i)
EA
=Q/ε₀
where Q(charge)=σA
EA=(σA)/ε₀
E=σ/ε₀
σ=ε₀E
σ=6.6375×10²nC/m²
For (b) total charge Q
Q=σA
There is attraction between the two poles. This is because like poles repel and unlike poles attract.
Hope it helped!
Answer:
Explanation:
The half life of a radioactive substance is the time in which the activity of the radioactive substance becomes half of its initial activity.
Or
The time in which the amount of radioactive substance becomes half of its initial amount in a sample.
It is denoted by T .
The relation between the decay constant and the half life is given by
Answer:
(D) Interference
Explanation:
Interference is a phenomenon characteristic for waves. When two waves meet, they interfere and result in a new wave. The precise shape of the resultant wave depends on the amplitude, frequency and phase of the constituent waves. In a most basic setting, two types of interference occur: constructive (resultant amplitude increases) and destructive (resultant amplitude decreases).
Answer:
The speed of the stone before it hit the river 3.00 sec later. Let v is the velocity at that instant is 45 m/s.
Explanation:
Given that, a child threw a stone straight down off a high bridge.
Initial velocity of the stone, u = 15 m/s
We need to find the speed of the stone before it hit the river 3.00 sec later. Let v is the velocity at that instant. When it come down, it is moving under the action of gravity. Using equation of motion as :
So, the speed of the stone before it hit the river 3.00 sec later. Let v is the velocity at that instant is 45 m/s.
