The work done by the electric field in moving a charge is the negative of the potential energy difference between the two locations, which is the product between the magnitude of the charge q and the potential difference
:
The proton charge is
, and the two locations have potential of
and
, therefore the work is
The work done on the car is -20 J.
Work done on the car is negative, meaning that the car actually does work on the external system.
<h3>Energy and law of conservation of energy</h3>
- Energy is the ability to do work
- the law of conservation of energy states that the total energy in a system is conserved
From the law of conservation of energy, the initial energy of the car before it moves down the road remains constant or unchanged.
- Initial energy = 100 J
- Initial energy = Final energy - work done on car
- Final Energy = Work done on car + initial energy
80J = Work done on car + 100 J
Work done on car = 80 - 100J
Work done on car = -20 J
Hence, the work done on the car is -20 J
Work done on car is negative.
Since work done on the car is negative, it means that the car actually does work on the external system. Hence, the decrease in the energy of the car.
Learn more about energy and work at: brainly.com/question/13387946
webpage of a scientist who is trying to sell a new invention
trying to dell = vested interest.
umiversity should be objective impartial
To solve this problem it is necessary to apply the law of Malus which describes the change in the Intensity of Light when it crosses a polarized surface.
Mathematically the expression is given as
Where,
= Initial Intensity
I = Final Intensity after pass through the polarizer
= Angle between the polarizer and the light
Since it is sought to reduce the intensity by half the relationship between the two intensities will be given as
Using the Malus Law we have,
Angle with respect to maximum is 
Q = C.v
v = Q/C
v = 4 × 10^(-10)/250
= 4 × 10^(-10)/2.5 × 10^2
= 1.6 × 10^(-12) volt
