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Cathode ray tubes (CRTs) used in old-style televisions have been replaced by modern LCD and LED screens.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 1.54 cm. If the potential difference across the plates was 27.0 kV, find the magnitude of the electric field (in V/m) in the region between the plates. HINT

Answer 1

Answer:

1753246.75325 V/m

Explanation:

d = Distance of separation = 1.54 cm

V = Potential difference = 27 kV

When the voltage is divided by the distance between the plates we get the electric field.

Electric field is given by

$E=\dfrac{V}{d}\\\Rightarrow E=\dfrac{27\times 10^3}{1.54\times 10^{-2}}\\\Rightarrow E=1753246.75325\ V/m$

The magnitude of the electric field in the region between the plates is 1753246.75325 V/m