Question

Two protons are launched with the same speed from point 1 inside the parallel-plate capacitor of FIGURE Q25.4. Points 2 and 3 are the same distance from the negative plate.

a. Is $\Delta U_{1\to 2}$, the change in potential energy along the path $1\to 2$, larger than, smaller than, or equal to $\Delta U_{1\to 3}$?

b. Is the proton's speed $v_2$at point 2 larger than, smaller than, or equal to $v_3$? Explain.

Short answer

a. Potential energy change along path $1\to 2$equals potential energy change along path $1\to 3$

b. Velocity at point 2 equals velocity at point 3.

Step-by-step solution

Given information

Motion of the two protons are

Along the path, there is a change in potential energy  and along the path 

(a)

The formula for the potential energy of a charge particle in an electric field is

$U=\frac{1}{4\pi {\Sigma }_v}\times \frac{q_{1}Q}{r}$

Here,

$U$is the potential energy

${\epsilon }_o$is electric permissibility

$q_1\&Q$are charges

$r$is the distance between charge particles

Because point 2 and point 3 are at the same vertical distance from the negative plate, the proton's potential energy at point 2 is the same as the proton's potential energy at point $3$.

As a result, the change in proton potential energy along path $1\to 2$ equals the change in proton potential energy along path $1\to 3$.

Points 2 and 3 have different proton speeds.

(b)

Since the change in potential energy along path $1\to 2$s the same as along path $1\to 3$, the change in kinetic energy is likewise the same, and the velocity of the proton being directly dependent on kinetic energy, the velocity of the proton is the same at both points 2 and 3.