Recall that there is a uniform electric field inside a capacitor. Recall also that in general, LaTeX: V=-\int{\vec{E}\cdot\vec{dl}}V = − ∫ E → ⋅ d l →. In regions of uniform field, this simplifies to LaTeX: V=|\vec{E}|dV = | E → | d, where d is the distance traveled parallel to the electric field. A capacitor with a gap of 0.5 mm has a potential difference from one plate to another of 22 V. What is the magnitude of the electric field within the capacitor? Give your answer in units of kV/m.

E = 44 kV / m

Explanation:

In a capacitor the potential difference is

         ΔV = -E x

         E = - ΔV / x

Let's reduce to the SI system

         x = 0.5 mm = 0.5 10⁻³ m

Let's calculate

         E = 22 / 0.5 10⁻³

         E = 44 10³ V / m

         E = 44 kV / m

Note that this electric field is uniform

beta is electron from nucleus

something from nucleus has changed, but charge must be same for cons of charge

when heated, the valance electrons increased their speed and transferred the heat from electron to electron.   i just took the test.