A proton is fired with a speed of 200,000m/s from the midpoint of the capacitor toward the positive plate. This speed is insufficient to reach the positive plate. What is the proton's speed as it collides with the negative plate

The proton speed = 3 × 10^5m/s

Explanation: The electric P.E change the proton if It can reach the positive plate.

The workdone

I have attached an image of the diagram showing the nature of this motion

Protons speed = 2.96 x 10^(5) m/s

Explanation:

A) At closest point of approach to the positive plate, the proton came to rest momentarily.

Thus;

Loss in Kinetic Energy = Gain in Electric potential energy

Hence;

(1/2)(mv^(2)) = eΔV

So, ΔV = (mv^(2))/(2e)

Mass of proton = 1.673 × 10-27 kilograms

Proton elementary charge(e) = 1.6 x 10^(-19) coulumbs

And from the question v = 200,000 m/s

So, ΔV = [1.673 × 10^(-27) x 200000^(2)] / (2 x 1.6 x 10^(-19)) = 209 V

This is less than 250V which is half of the charge at the positive plate shown in the diagram.

Therefore, the speed is insufficient to reach the positive plate from P to Q.

B) Gain in KE = qΔV

Thus; 1/2mvf^(2) - 1/2mvi^(2) = eΔV

Where, vf is final velocity and vi is initial velocity.

So simplifying, we get;

vf^(2) - vi^(2) = (2eΔV)/m

So, vf = √[(2eΔV)/m) + (vi^(2))

= √[(2 x 1.6 x 10^(-19) x 250)/(1.673 × 10^(-27)) + (200,000^(2))

= 2.96 x 10^(5) m/s

weight of the box is given as

here this is placed on the inclined of angle 42 degree

now this weight will have two components one is along the incline other is perpendicular to it

so here we need to apply a force parallel to incline opposite to its component of weight

so here the component of weight is given as

here we have

so it requires 75 n force to move it upwards

305.7

explanation:

density =m/v

density =917/3.00

density = 305.

approximately = 305.7

in four significant figures