The expression for the magnitude of the electric field between two uniform conducting plates is
Here, V is potential difference between plates and d is separation between plates.
As the potential 6.00 cm from the zero volt plate (and 4.00 cm from the other) is 420 V.
Therefore,
Thus, the electric field strength between the plates is 7000 V/ m
Answer:
Explanation:
The speed increased from 2.0 * 10^7 m/s to 4.0 * 10^7 m/s over a 1.2 cm distance.
Let us find the acceleration:
Electric force is given as the product of charge and electric field strength:
F = qE
where q = electric charge
E = Electric field strength
Force is generally given as:
F = ma
where m = mass
a = acceleration
Equating both:
ma = qE
E = ma / q
For an electron:
m = 9.11 × 10^{-31} kg
q = 1.602 × 10^{-19} C
Therefore, the electric field strength of the electron is: