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3 years ago

Two parallel plates have equal but opposite charges on their surface. The plates are separated by a finite distance.

Physics

1 answer:

miss Akunina [59]3 years ago

6 0

Answer:

(a) The speed of proton decreases as it moves from A to B.

(b) Plate B is at a higher potential.

(d) Parallel lines parallel to the two plates.

Parallel lines equally spaced.

Explanation:

The electric potential energy is given by the following formula:

$U = \frac{1}{4\pi \epsilon_0}\frac{qq_0}{r}$

Alternatively, potential energy in a uniform electric field is

$U = qEr$

where 'r' is the distance from negative to positive plates. This definition is analogues to that of gravitational potential energy, U = mgh.

If the positively charged proton is gaining potential energy as it gets closer to plate B, then plate B is charged positively.

(a) According to this information, the speed of proton decreases as it moves from A to B. This is similar to the speed of an object which is gaining potential energy by moving higher.

(b) By the same gravitational analogy, plate B is at a higher potential.

(d) The equipotential lines are parallel to electric field lines which are perpendicular to the plates. So, the equipotential lines are parallel to the plates. Since the electric field between the plates is uniform, then the equipotential lines are equally seperated.

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a) The exit velocity of the projectile is 16.7 m/s

b) The maximum height achieved by the projectile is 14.2 m

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We solve this first part of the problem by applying the work-energy theorem, which states that the work done on the projectile is equal to the gain in kinetic energy of the projectile. Mathematically:

$W=Fd = \frac{1}{2}mv^2-\frac{1}{2}mu^2=\Delta K$

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F = 150 N is the force applied

d = 1.25 m is the displacement of the projectile (the length of the firing arm)

m = 1350 g = 1.35 kg is the mass of the projectile

u = 0 is the initial velocity of the projectile

v is the exit velocity of the projectile

Solving for v, we find:

$v=\sqrt{\frac{2Fd}{m}}=\sqrt{\frac{2(150)(1.25)}{1.35}}=16.7 m/s$

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$K_i = U_f$

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where:

$K_i$ is the initial kinetic energy

$U_f$ is the final potential energy

m = 1350 g = 1.35 kg is the mass of the projectile

v = 16.7 m/s is the velocity at which the projectile leaves the cannon

$g=9.8 m/s^2$ is the acceleration of gravity

h is the maximum height reached by the projectile

And solving for h, we find

$h=\frac{v^2}{2g}=\frac{(16.7)^2}{2(9.8)}=14.2 m$

Assuming the projectile is launched vertically upward, then the total time of flight is twice the time it takes for reaching the maximum height. This time can be found by using the following suvat equation:

$v=u-gt$