Answer:
The total electric potential at mid way due to 'q' is ![\frac{q}{4\pi\epsilon_{o}d}](https://tex.z-dn.net/?f=%5Cfrac%7Bq%7D%7B4%5Cpi%5Cepsilon_%7Bo%7Dd%7D)
The net Electric field at midway due to 'q' is 0.
Solution:
According to the question, the separation between two parallel plates, plate A and plate B (say) = d
The electric potential at a distance d due to 'Q' is:
![V = \frac{1}{4\pi\epsilon_{o}}.\frac{Q}{d}](https://tex.z-dn.net/?f=V%20%3D%20%5Cfrac%7B1%7D%7B4%5Cpi%5Cepsilon_%7Bo%7D%7D.%5Cfrac%7BQ%7D%7Bd%7D)
Now, for the Electric potential for the two plates A and B at midway between the plates due to 'q':
For plate A,
Similar is the case with plate B:
Since the electric potential is a scalar quantity, the net or total potential is given as the sum of the potential for the two plates:
![V_{total} = V_{A} + V_{B} = \frac{1}{4\pi\epsilon_{o}}.q(\frac{1}{\frac{d}{2}} + \frac{1}{\frac{d}{2}}](https://tex.z-dn.net/?f=V_%7Btotal%7D%20%3D%20V_%7BA%7D%20%2B%20V_%7BB%7D%20%3D%20%5Cfrac%7B1%7D%7B4%5Cpi%5Cepsilon_%7Bo%7D%7D.q%28%5Cfrac%7B1%7D%7B%5Cfrac%7Bd%7D%7B2%7D%7D%20%2B%20%5Cfrac%7B1%7D%7B%5Cfrac%7Bd%7D%7B2%7D%7D)
![V_{total} = \frac{q}{4\pi\epsilon_{o}d}](https://tex.z-dn.net/?f=V_%7Btotal%7D%20%3D%20%5Cfrac%7Bq%7D%7B4%5Cpi%5Cepsilon_%7Bo%7Dd%7D)
Now,
The Electric field due to charge Q at a distance is given by:
![\vec{E} = \frac{1}{4\pi\epsilon_{o}}.\frac{Q}{d^{2}}](https://tex.z-dn.net/?f=%5Cvec%7BE%7D%20%3D%20%5Cfrac%7B1%7D%7B4%5Cpi%5Cepsilon_%7Bo%7D%7D.%5Cfrac%7BQ%7D%7Bd%5E%7B2%7D%7D)
Now, if the charge q is mid way between the field, then distance is
.
Electric Field at plate A,
at midway due to charge q:
![\vec{E_{A}} = \frac{1}{4\pi\epsilon_{o}}.\frac{q}{(\frac{d}{2})^{2}}](https://tex.z-dn.net/?f=%5Cvec%7BE_%7BA%7D%7D%20%3D%20%5Cfrac%7B1%7D%7B4%5Cpi%5Cepsilon_%7Bo%7D%7D.%5Cfrac%7Bq%7D%7B%28%5Cfrac%7Bd%7D%7B2%7D%29%5E%7B2%7D%7D)
Similarly, for plate B:
![\vec{E_{B}} = \frac{1}{4\pi\epsilon_{o}}.\frac{q}{(\frac{d}{2})^{2}}](https://tex.z-dn.net/?f=%5Cvec%7BE_%7BB%7D%7D%20%3D%20%5Cfrac%7B1%7D%7B4%5Cpi%5Cepsilon_%7Bo%7D%7D.%5Cfrac%7Bq%7D%7B%28%5Cfrac%7Bd%7D%7B2%7D%29%5E%7B2%7D%7D)
Both the fields for plate A and B are due to charge 'q' and as such will be equal in magnitude with direction of fields opposite to each other and hence cancels out making net Electric field zero.