A parallel-plate capacitor has plates with an area of 451 cm2 and an air-filled gap between the plates that is 2.51 mm thick. Th
1 answer:
To solve this problem we will apply the concepts related to Energy defined in the capacitors, as well as the capacitance and load. From these three definitions we will build the solution to the problem by defending the energy with the initial conditions, the energy under new conditions and finally the change in the work done to move from one point to the other.
Energy in a capacitor can be defined as
Here,
V = Potential difference across the capacitor plates
Q = Charge stored on the capacitor plates
At the same time capacitance can be defined as,
Here,
Vacuum permittivity constant
A = Area
d = Distance
Replacing with our values we have that,
PART A) Energy stored in the capacitor is
PART B) We know first that everything that the load can be defined as the product between voltage and capacitance, therefore
Now if we have that the capacitance is
Then the energy stored is
PART C) The amount of work or energy required to carry out this process is the difference between the energies obtained, therefore
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1) The motion of the ball is an uniformly accelerated motion, with constant acceleration equal to 
(the negative sign means it is directed towards the ground).
For an uniformly accelerated motion, we can use the following relationship:
where h is the maximum height reached by the ball,
is its initial velocity and 
the velocity of the ball when it reaches the maximum height. But 
(when the ball reaches the maximum height, it stops before going down, so its velocity at that moment is zero), so we can use the relationship to calculate h, the maximum height:
2) We can find the time the ball takes to return to the ground by requiring that the space covered by the ball returns to zero:
where for an uniformly accelerated motion,
By solving this, we have two solutions: one is t=0, which corresponds to the moment the player hits the ball, the second one is
so, the ball returns to the ground after 4 s.
3) The velocity of the ball when it returns to the ground is given by:
where the negative sign means the ball is going downwards.
For an uniformly accelerated motion, we can use the following relationship:
where h is the maximum height reached by the ball,
2) We can find the time the ball takes to return to the ground by requiring that the space covered by the ball returns to zero:
where for an uniformly accelerated motion,
By solving this, we have two solutions: one is t=0, which corresponds to the moment the player hits the ball, the second one is
so, the ball returns to the ground after 4 s.
3) The velocity of the ball when it returns to the ground is given by:
where the negative sign means the ball is going downwards.
The change in surface area of Gaussian surface with radius (r) is 8πr.
<h3>Electric field from Coulomb's law</h3>The electric field experienced by a charge is calculated as follows;
where;
- E is the electric field
- Q is the charge
- r is the radius
The electric field reduces by a factor of
The surface area of a sphere is given as;
Thus, the change in surface area of Gaussian surface with radius (r) is 8πr.
Learn more about area of Gaussian surfaces here: brainly.com/question/17060446