Question

Cathode ray tubes (CRTs) used in old-style televisions have been replaced by modern LCD and LED screens. Part of the CRT included a set of accelerating plates separated by a distance of about 1.40 cm. If the potential difference across the plates was 23.0 kV, find the magnitude of the electric field (in V/m) in the region between the plates.

Answer 1

Answer:

E =  1.64 x 10⁶ V/m

Explanation:

The electric field in the region between the plates can be given by the following formula:

$E = \frac{\Delta V}{d}$

where,

E = Electric Field = ?

ΔV = Poetential Difference across the plates = 23 KV = 23000 V

d = distance between plates = 1.4 cm = 0.014 m

Therefore, using these values in the equation, we get:

$E = \frac{23000\ V}{0.014\ m}$

E =  1.64 x 10⁶ V/m