# Find the magnitude of the electric field in the capacitor

• titi07
In summary, the conversation discusses finding the magnitude of the electric field and the speed of an electron in a parallel-plate capacitor. The electron enters the capacitor with a speed of v=5.20 x 10^-6 m/s and is deflected downward by a distance of d=0.594cm at the point of exit. To solve this problem, one can use the electric force on a charge in an electric field and apply it to Newton's second law. Additionally, kinematic equations, such as those used in projectile motion, can be utilized.
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

## The Attempt at a Solution

I have no clue how to start this problem

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).

.

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.

## 1. What is the definition of an electric field in a capacitor?

The electric field in a capacitor is a measure of the force that an electric charge experiences within the capacitor. It is defined as the force per unit charge and is represented by the symbol E.

## 2. How do you calculate the magnitude of the electric field in a capacitor?

The magnitude of the electric field in a capacitor can be calculated by dividing the voltage across the capacitor by the distance between the plates. This can be expressed as E = V/d, where E is the electric field, V is the voltage, and d is the distance between the plates.

## 3. What is the unit of measurement for the magnitude of the electric field in a capacitor?

The unit of measurement for the magnitude of the electric field in a capacitor is volts per meter (V/m).

## 4. How does the distance between the plates affect the magnitude of the electric field in a capacitor?

The magnitude of the electric field in a capacitor is directly proportional to the distance between the plates. This means that as the distance between the plates increases, the electric field decreases and vice versa.

## 5. What factors can affect the magnitude of the electric field in a capacitor?

The magnitude of the electric field in a capacitor can be affected by the voltage applied, the distance between the plates, the material and shape of the plates, and the dielectric material between the plates. Changes in any of these factors can alter the electric field in the capacitor.

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