Question

The electron gun in an old TV picture tube accelerates electrons between two parallel plates 1.5 cm apart with a 30 kV potential difference between them. The electrons enter through a small hole in the negative plate, accelerate, then exit through a small hole in the positive plate. Assume that the holes are small enough not to affect the electric field or potential. a)What is the electric field strength between the plates?

Answer 1

The electric field strength between the plates of an electron gun with a potential difference of 30 kV and plate separation of 1.5 cm is 2,000,000 V/m.

Step-by-step explanation:

To calculate the electric field strength between the plates of an electron gun, you can use the relationship E = V/d, where E is the electric field strength, V is the potential difference, and d is the separation between the plates.

In this case, the potential difference V is 30 kilovolts (30,000 volts), and the separation d is 1.5 cm (or 0.015 meters). Thus, the electric field strength E between the plates can be calculated as follows:

E = V/d = 30,000 volts / 0.015 meters = 2,000,000 V/m

The electric field strength is therefore 2,000,000 volts per meter (V/m).

Answer 2

For a pair of charged parallel plates (think capacitor) where the plate size is much bigger compared to the plate spacing, we can approximate the electric field as having a constant magnitude everywhere between the plates. The electric field strength is given by:

E = ΔV/d

E = electric field strength, ΔV = potential difference, d = plate spacing

Given values:

ΔV = 30×10³V, d = 1.5×10⁻²m

Plug in and solve for E:

E = 2 megavolts/meter