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3 years ago

"In the diagram below, point P is located in the electric field between two oppositely charged parallel plates.

Physics

2 answers:

egoroff_w [7]3 years ago

6 0

Explanation :

From the given figure, point P is located in the electric field between two oppositely charged parallel plates.

If an electron is placed at point P, it will repel by the negative plate of the capacitor and will move towards positive plate i.e. from the negative plate to positive plate.

Similarly, when a proton is placed at point P, it will attract towards the negative plate i.e. the direction of the proton will be from positive plate to negative plate.

We know that the electrostatic force is inversely proportional to the square of the distance.

So, the electrostatic force on a proton placed at point P has a greater magnitude and the opposite direction".

olya-2409 [2.1K]3 years ago

5 0

As proton is heavier then electron, and electrostatic force is directly proportional to mass, so it would be greater and direction would be changed 'cause of different charges.

In short, Your Answer would be Option 4

Hope this helps!

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A charge of -3.02 μC is fixed in place. From a horizontal distance of 0.0377 m, a particle of mass 9.43 x 10^-3 kg and charge -9

Andreyy89

Answer:

$d = 0.0306 m$

Explanation:

Here we know that for the given system of charge we have no loss of energy as there is no friction force on it

So we will have

$U + K = constant$

$\frac{kq_1q_2}{r_1} + \frac{1}{2}mv_1^2 = \frac{kq_1q_2}{r_2} + \frac{1}{2}mv_2^2$

now we know when particle will reach the closest distance then due to electrostatic repulsion the speed will become zero.

So we have

$\frac{(9 \times 10^9)(3.02 \mu C)(9.78 \mu C)}{0.0377} + \frac{1}{2}(9.43 \times 10^{-3})(80.4)^2 = \frac{(9 \times 10^9)(3.02 \mu C)(9.78 \mu C)}{r} + 0$

$7.05 + 30.5 = \frac{0.266}{r}$

$r = 7.08 \times 10^{-3} m$

so distance moved by the particle is given as

$d = r_1 - r_2$

$d = 0.0377 - 0.00708$

$d = 0.0306 m$

6 0

2 years ago

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dmitriy555 [2]

Answer:

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2 years ago

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The three forces acting on a hot-air balloon that is moving vertically are its weight, the force due to air resistance and the u

irinina [24]

Explanation :

The forces acting on hot- air balloon are:

Weight, (W)

Force due to air resistance, (F)

Upthrust force, (U)

Its weight W is acting in downward direction. The upthrust force U acts in upward direction. When the balloon is moving upward, the air resistance is in downward and vice versa.

In this case, the hot-air balloon descends vertically at constant speed.

so, $a=0$