Electric Field Required to Accelerate Electrons in Television Tube

[pringless]

The electron gun in a televeision tube is used to accelerate electrons (mass of 9.10939 x 10^-31 kg and charge of -1.60218 x 10^-19 C) from rest to 2 x 10^7 m/s within a distance of .053 m. What electric field is required?

 

[ShawnD]

The question only looks complicated because of the big (or small) numbers. You are given a change in velocity, a distance, and a charge.

First find the acceleration on the electron with this equation.

$$V_f^2 = V_i^2 + 2ad$$

Now find the force on the electron with this equation

$$F = ma$$

Now find the electric field with this equation

$$E = \frac{F}{q}$$

 

[M98Ranger]

To calculate the required electric field, we can use the equation F = qE, where F is the force exerted on the electron, q is the charge of the electron, and E is the electric field.

First, we need to calculate the force required to accelerate the electron to a speed of 2 x 10^7 m/s. We can use the equation F = ma, where m is the mass of the electron and a is the acceleration. Plugging in the given values, we get F = (9.10939 x 10^-31 kg)(2 x 10^7 m/s^2) = 1.821878 x 10^-23 N.

Next, we can rearrange the equation F = qE to solve for E. This gives us E = F/q. Plugging in the values for F and q, we get E = (1.821878 x 10^-23 N)/(-1.60218 x 10^-19 C) = -1.136 x 10^5 N/C.

Therefore, the required electric field to accelerate the electrons in the television tube is -1.136 x 10^5 N/C. The negative sign indicates that the electric field must be directed towards the electron gun, in the opposite direction of the electron's motion, in order to accelerate the electrons.