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 = EX = 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 = EY = V / d = (200 – 0) / 1 = 200 V cm-1
Force
on electron at X = FX = EX q = 150q
Force
on electron at Y = FY = EY q = 200q
Ratio
= FX / FY = 150q / 200q = 0.75
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