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OleMash [197]
3 years ago
13

There are two parallel conductive plates separated by a distance d and zero potential. Calculate the potential and electric fiel

d that occurs if a q charge is placed between the plates at a distance d/2.
Physics
1 answer:
taurus [48]3 years ago
3 0

Answer:

The total electric potential at mid way due to 'q' is \frac{q}{4\pi\epsilon_{o}d}

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}

Now, for the Electric potential for the two plates A and B at midway between the plates due to 'q':

For plate A,

V_{A} = \frac{1}{4\pi\epsilon_{o}}.\frac{q}{\frac{d}{2}}

Similar is the case with plate B:

V_{B} = \frac{1}{4\pi\epsilon_{o}}.\frac{q}{\frac{d}{2}}

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}}

V_{total} = \frac{q}{4\pi\epsilon_{o}d}

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}}

Now, if the charge q is mid way between the field, then distance is \frac{d}{2}.

Electric Field at plate A, \vec{E_{A}} at midway due to charge q:

\vec{E_{A}} = \frac{1}{4\pi\epsilon_{o}}.\frac{q}{(\frac{d}{2})^{2}}

Similarly, for plate B:

\vec{E_{B}} = \frac{1}{4\pi\epsilon_{o}}.\frac{q}{(\frac{d}{2})^{2}}

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.

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