Answer:
-30 N/C
Explanation:
Since the potential changes from 0.90 V to 1.2 V when I move the probe 1 cm closer to the non-grounded electrode, the electric field is the gradient between the two points is given by E = -ΔV/Δx where ΔV = change in electric potential and Δx = distance of potential change = 1 cm = 0.01 m
Now ΔV = final potential - initial potential = 1.2 V - 0.90 V = 0.30 V
Since E = -ΔV/Δx
substituting the values of the variables into the equation, we have
E = -ΔV/Δx
E = -0.30 V/0.01 m
E = -30 V/m
Since 1 V/m = 1 N/C.
E = -30 N/C
So, the average electric field is -30 N/C
Answer:
No
Explanation:
20 billion light-years away are beyond our sight and perspective on Earth and wouldn't be observable in our universe.
Answer:
Explanation:
From the question we are told that:
Frictional force 
Coefficient of kinetic friction 
Generally the equation for Normal for is mathematically given by
Therefore
Answer:
How far will the electron travel beforehitting a plate is 248.125mm
Explanation:
Applying Gauss' law:
Electric Field E = Charge density/epsilon nought
Where charge density=1.0 x 10^-6C/m2 & epsilon nought= 8.85× 10^-12
Therefore E = 1.0 x 10^-6/8.85× 10^-12
E= 1.13×10^5N/C
Force on electron F=qE
Where q=charge of electron=1.6×10^-19C
Therefore F=1.6×10^-19×1.13×10^5
F=1.808×10^-14N
Acceleration on electron a = Force/Mass
Where Mass of electron = 9.10938356 × 10^-31
Therefore a= 1.808×10^-14 /9.11 × 10-31
a= 1.985×10^16m/s^2
Time spent between plate = Distance/Speed
From the question: Distance=1cm=0.01m and speed = 2×10^6m/s^2
Therefore Time = 0.01/2×10^6
Time =5×10^-9s
How far the electron would travel S =ut+ at^2/2 where u=0
S= 1.985×10^16×(5×10^-9)^2/2
S=24.8125×10^-2m
S=248.125mm
Answer:
2.1km
Explanation:
Ill take it as u are talking about the displacement
Since displacement has negatives and positves
5.9 - 3.8 = 2.1km
