centre of square disrance to each corner found by Pythagoras' theorem.
coulombs law used to clculate field of each charge at centre
fields added vectorially for res
Answer:
The pushing of the car by you and your friends is the applied force and when the car moves, it means that the velocity has changed thus causing the movement or acceleration.
Explanation:
Newton's work on forces regarding motion can never be neglected by scientists. Sir Isaac Newton when he was alive, among several of his works he proposed the three laws guiding the forces of motion. In this question we are only going to be treating only one out of the three Newton's Law of motion and that is the second Law Of Newton's laws of motion.
The second Law Of Newton's laws of motion states that the acceleration of an object is directly proportional to the applied force and inversely proportional to the object's mass.
(1). Now, to the question: " How are you and your friends applying Newton's second law of motion here? "
The pushing of the car by you and your friends is the applied force and when the car moves, it means that the velocity has changed thus causing the movement or acceleration.
According to the law, the more the Force, the more the acceleration.
(2). For the second part of the question, " What if the car you were traveling in was a large SUV?"
From the law stated above we see that the acceleration is inversely proportional to the mass, thus if the car is a large SUV, It means that more force is needed to change the car's velocity.
Complete question:
A uniform electric field is created by two parallel plates separated by a
distance of 0.04 m. What is the magnitude of the electric field established
between the plates if the potential of the first plate is +40V and the second
one is -40V?
Answer:
The magnitude of the electric field established between the plates is 2,000 V/m
Explanation:
Given;
distance between two parallel plates, d = 0.04 m
potential between first and second plate, = +40V and -40V respectively
The magnitude of the electric field established between the plates is calculated as;
E = ΔV / d
where;
ΔV is change in potential between two parallel plates;
d is the distance between the plates
ΔV = V₁ -V₂
ΔV = 40 - (-40)
ΔV = 40 + 40
ΔV = 80 V
E = ΔV / d
E = 80 / 0.04
E = 2,000 V/m
Therefore, the magnitude of the electric field established between the plates is 2,000 V/m
Answer:
E = 5291.00 N/C
Explanation:
Expression for capacitance is
where
A is area of square plate
D = DISTANCE BETWEEN THE PLATE
We know that capacitrnce and charge is related as
v = 9.523 V
Electric field is given as
= 
E = 5291.00 V/m
E = 5291.00 N/C
Electric potential energy is defined as Ep=Q*V where Q is the magnitude of the charge and V is the potential difference. So when a charge moves between the points that have a potential difference, it's energy changes.
In our case:
Q=2e=2*(-1.6*10^-19) C
V=75 V
Ep=(-3.2*10^-19)*75
Ep=-2.4*10^-17 J
The change in potential energy of the charge is -2.4*10^-17 J
