The magnitude of the electric field at the dot is : 10⁴ v/m
<h3>What is electric field?</h3>
Electric field is a region around a charged particle or object within which a force would be exerted on other charged particles or objects.
Given that there are three equipotential lines with equal spacing,we will apply the the relationship between P.D and electric field
Determine the magnitude of the electric field at the dot
change in voltage = E .d
From equation ( 1 )
The magnitude of electric field will be given as
Hence we can conclude that The magnitude of the electric field at the dot is : 10⁴ v/m
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Answer:8.1 m
Explanation:
Given
ball is launched from height of 3 m
initial velocity 
considering the ball is thrown vertically upward
Using 
where,
u=initial velocity
v=final Velocity
a=acceleration
s=distance
At maximum height final velocity will be zero
Therefore maximum height w.r.t ground is 
Coulomb's Law
Given:
F = 3.0 x 10^-3 Newton
d = 6.0 x 10^2 meters
Q1 = 3.3x 10^-8 Coulombs
k = 9.0 x 10^9 Newton*m^2/Coulombs^2
Required:
Q2 =?
Formula:
F = k • Q1 • Q2 / d²
Solution:
So, to solve for Q2
Q2 = F • d²/ k • Q1
Q2 = (3.0 x 10^-3 Newton) • (6.0 x 10^2 m)² / (9.0 x 10^9
Newton*m²/Coulombs²) • (3.3x 10^-8 Coulombs)
Q2 = (3.0 x 10^-3 Newton) • (360 000 m²) / (297 Newton*m²/Coulombs)
Q2 = 1080 Newton*m²/ (297 Newton*m²/Coulombs)
Then, take the reciprocal of the denominator and start
multiplying
Q2 = 1080 • 1 Coulombs/297
Q2 = 1080 Coulombs / 297
Q2 = 3.63636363636 Coulombs
Q2 = 3.64 Coulumbs
Explanation:
Work is the dot product of the force and displacement vectors.
W = F · d
In other words, it is the force times the parallel component of the distance.
W = F d cos θ, where θ is the angle between the force and distance.
Answer:
31.25 m
25m/sec
Explanation:
Given :-
Time = 5sec
V = 0 (in going up)
U = 0 (in comming down)
Find :-
H and U by which it is thrown up
Since the total time is 5 sec ,therefore half time will be taken to go up and another half will be taken to go down .
We know that ,
V = U + gt
0 = U - 10*2.5
U = 25 m/sec
Also,
V² = U² +2gs
0 = 625 - 20s
s = 625/20 = 31.25 m
