Respuesta :
Answer:
The total electric potential at mid way due to 'q' is [tex]\frac{q}{4\pi\epsilon_{o}d}[/tex]
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:
[tex]V = \frac{1}{4\pi\epsilon_{o}}.\frac{Q}{d}[/tex]
Now, for the Electric potential for the two plates A and B at midway between the plates due to 'q':
For plate A,
[tex]V_{A} = \frac{1}{4\pi\epsilon_{o}}.\frac{q}{\frac{d}{2}}[/tex]
Similar is the case with plate B:
[tex]V_{B} = \frac{1}{4\pi\epsilon_{o}}.\frac{q}{\frac{d}{2}}[/tex]
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:
[tex]V_{total} = V_{A} + V_{B} = \frac{1}{4\pi\epsilon_{o}}.q(\frac{1}{\frac{d}{2}} + \frac{1}{\frac{d}{2}}[/tex]
[tex]V_{total} = \frac{q}{4\pi\epsilon_{o}d}[/tex]
Now,
The Electric field due to charge Q at a distance is given by:
[tex]\vec{E} = \frac{1}{4\pi\epsilon_{o}}.\frac{Q}{d^{2}}[/tex]
Now, if the charge q is mid way between the field, then distance is [tex]\frac{d}{2}[/tex].
Electric Field at plate A, [tex]\vec{E_{A}}[/tex] at midway due to charge q:
[tex]\vec{E_{A}} = \frac{1}{4\pi\epsilon_{o}}.\frac{q}{(\frac{d}{2})^{2}}[/tex]
Similarly, for plate B:
[tex]\vec{E_{B}} = \frac{1}{4\pi\epsilon_{o}}.\frac{q}{(\frac{d}{2})^{2}}[/tex]
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.
