A pair of oppositely charged parallel plates is
separated by 5.44 mm. A potential difference
of 626 V exists between the plates.
What is the strength of the electric field
between the plates? The fundamental charge
is 1.602 × 10−19. What is the magnitude of the force on an electron between the plates?

Question

A pair of oppositely charged parallel plates is
separated by 5.44 mm. A potential difference
of 626 V exists between the plates.
What is the strength of the electric field
between the plates? The fundamental charge
is 1.602 × 10−19. What is the magnitude of the force on an electron between the plates?

anonymous · Answer

$$E=V/d=(626)/(5.44x10^{-3})=1.15x10^{6} N/C$$

anonymous · Answer

$$F=E.q=(1.15x10^{6}).(1.602x10^{-19})=1.84x10^{-14}N$$

anonymous · Answer

How much work must be done on the electron
to move it to the negative plate if it is initially
positioned 2.74 mm from the positive plate?