Three parallel metal plates of the same area are fixed with a separation of 2.0 cm between the top plate and the centre plate, and 1.0 cm between the centre plate and the bottom plate.

Question 4

Three parallel metal plates of the same area are fixed with a separation of 2.0 cm between the top plate and the centre plate, and 1.0 cm between the centre plate and the bottom plate. The top plate is held at a potential of +500 V, the middle plate at +200 V and the bottom plate is earthed, as shown.

What is the value of the ratio

magnitude of force on an electron at X / magnitude of force on an electron at Y ?

A 0.75              B 1.00              C 1.25             D 1.50

Reference: Past Exam Paper – November 2015 Paper 11 Q30

Solution:

Answer: A.

Electric force = Eq      where q is the charge of the electron

Electric field = V / d   where V is the p.d. between the plates and d is the separation of the plates

The electric field strength depends on the potential difference between the plates. It is NOT necessarily the potential of the top plate.

X is found between the top plate (at + 500 V) and the centre plate (at + 200 V).

Electric field strength at X = E_X = V / d = (500 – 200) / 2 = 150 V cm^-1

Y is found between the centre plate (at +200 V) and the bottom plate (at 0 V).

Electric field strength at Y = E_Y = V / d = (200 – 0) / 1 = 200 V cm^-1

Force on electron at X = F_X = E_X q = 150q

Force on electron at Y = F_Y = E_Y q = 200q

Ratio = F_X / F_Y = 150q / 200q = 0.75