Two parallel plates having charges of equal magnitude but opposite sign are separated by 11.0 cm. Each plate has a surface charge density of 49.0 nC/m2. A proton is released from rest at the positive plate. (a) Determine the magnitude of the electric field between the plates from the charge density.

Two parallel plates having charges of equal magnitude but opposite sign are separated by 11.0 cm. Each plate has a surface charge density of 49.0 nC/m2. A proton is released from rest at the positive plate.

(a) Determine the magnitude of the electric field between the plates from the charge density.

b) Determine the potential difference between the plates.

a

The Electric Field between the two plate is  

b

The potential difference between the plate is V =

Explanation:

From the we are given that

        The separation between the plate is  

        The surface charge density is  

Generally Electric field between the plate is mathematically given as

Note that is the permitivity of free space and its value is  

    Now substituting values we have

Generally Potential difference between the plate is mathematically given as

Where E is the electric field  which is  

                              Substituting value we have

                           V =                  

nah.   i'm just sleepy.