Between two large, flat, oppositely charged plates, the electric field is the same everywhere: the same size and the same direction. A charge there feels a constant force, so it moves just like a ball in projectile motion, with the electric force playing the part of gravity.
Why it works
Each plate is a sheet of charge whose field does not weaken with distance (for points close to a large sheet). Between the plates the two fields point the same way and add; outside they point opposite ways and cancel. So the field is uniform between the plates and nearly zero outside. A uniform field gives a constant force $qE$ and a constant acceleration $a = qE/m$, and all of constant-acceleration kinematics applies.
Key idea
In a uniform field, $a = \dfrac{qE}{m}$, constant. A positive charge accelerates along $\vec E$; an electron accelerates opposite to $\vec E$. A charge entering sideways follows a parabola: constant speed along the plates, constant acceleration across them.
Worked examples
Worked example: A proton's acceleration
Find the acceleration of a proton in a uniform field of $2.0 \times 10^{4}\,\text{N/C}$.
- $$a = \dfrac{eE}{m_p} = \dfrac{(1.60 \times 10^{-19})(2.0 \times 10^{4})}{1.67 \times 10^{-27}} = 1.92 \times 10^{12}\,\text{m/s}^2.$$
- About $10^{11}$ times $g$, which is why gravity is ignored here.
Answer: $1.92 \times 10^{12}\,\text{m/s}^2$, along the field
Worked example: Bending an electron beam
An electron enters the gap between plates $4.0\,\text{cm}$ long at $2.0 \times 10^{6}\,\text{m/s}$, moving parallel to them. The field is $500\,\text{N/C}$. How far sideways is it pushed?
- Time between the plates: $t = \dfrac{0.040\,\text{m}}{2.0 \times 10^{6}\,\text{m/s}} = 2.0 \times 10^{-8}\,\text{s}$.
- Sideways acceleration: $a = \dfrac{(1.60 \times 10^{-19})(500)}{9.11 \times 10^{-31}} = 8.78 \times 10^{13}\,\text{m/s}^2$.
- $$y = \tfrac{1}{2}at^2 = \tfrac{1}{2}(8.78 \times 10^{13})(2.0 \times 10^{-8})^2 = 0.0176\,\text{m} = 1.76\,\text{cm},$$ toward the positive plate.
Answer: $1.76\,\text{cm}$
Worked example: A floating oil drop
An oil drop of mass $3.27 \times 10^{-15}\,\text{kg}$ hangs at rest in a downward field of $1.0 \times 10^{5}\,\text{N/C}$. What is its charge?
- At rest: the electric force balances the weight, $|q|E = mg$, so $|q| = \dfrac{(3.27 \times 10^{-15})(9.8)}{1.0 \times 10^{5}} = 3.20 \times 10^{-19}\,\text{C}$.
- That is $2e$. The force must point up, against the downward field, so the drop is negative: $q = -2e$. Millikan found charges like this were always whole multiples of $e$.
Answer: $-3.20 \times 10^{-19}\,\text{C} = -2e$
Common mistake
Forgetting that an electron is pushed opposite to the field. $\vec F = q\vec E$, and for an electron $q$ is negative. In a field pointing up, an electron accelerates down. Check: an electron always moves toward the positive plate.