Find the magnitude of the electric field in the capacitor

[titi07]

Homework Statement

An electron entering a parallel-plate capacitor with a speed of v=5.20 x 10^-6 m/s. The electric field of the capacitor has deflected the electron downward by a distance of d=0.594cm at the point where the electron exists the capacitor.
a)find the magnitude of the electric field in the capacitor____N/C
b)find the speed of the electron when it exists the capacitor___m/s

Homework Equations

The Attempt at a Solution

I have no clue how to start this problem

 

[hage567]

What's the electric force on a charge in an electric field? Apply this to Newton's second law.

Then think about your kinematic equations (like in projectile motion).

 

[Moetasim]

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I would begin by first identifying the relevant equations and principles that can help solve this problem. In this case, the relevant equation is the equation for the force experienced by a charged particle in an electric field: F = qE, where F is the force, q is the charge of the particle, and E is the electric field.

Next, I would gather the given information: the speed of the electron (v = 5.20 x 10^-6 m/s), the distance it is deflected (d = 0.594 cm = 0.00594 m), and the charge of an electron (q = -1.6 x 10^-19 C).

To solve for the magnitude of the electric field (E), I would rearrange the equation to solve for E: E = F/q. Since the electron is deflected downwards, the direction of the force must also be downwards. Therefore, the magnitude of the force can be calculated using the equation F = ma, where m is the mass of the electron and a is the acceleration due to the electric field. The acceleration can be calculated using the equation a = v^2/d.

Plugging in the values, we get: a = (5.20 x 10^-6 m/s)^2 / 0.00594 m = 8.76 x 10^-9 m/s^2. And since F = ma, the force is: F = (9.11 x 10^-31 kg)(8.76 x 10^-9 m/s^2) = 7.98 x 10^-39 N.

Finally, plugging in the values for q and F into the equation E = F/q, we get: E = (7.98 x 10^-39 N)/(-1.6 x 10^-19 C) = -4.99 x 10^-20 N/C.

To find the speed of the electron when it exits the capacitor, we can use the equation v = sqrt(2qV/m), where V is the potential difference between the plates of the capacitor and m is the mass of the electron. Since the electron is moving from a higher potential (inside the capacitor) to a lower potential (outside the capacitor), V is negative. We can calculate V using the equation V = Ed, where E is the electric field we just calculated and d is the distance the electron is deflected.