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I have this problem in my physics homework, which is not trivial. This is not a graded homework fortunately but I would like to see the solution for it and learn. I have attached my attempt at solving this problem. Not sure if this is the right way to solve it, and so I would like your personal input in this one.

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Problem:

If a charge is fixed above a large, flat, grounded, conducting floor, the electric field above the floor would be the same as that produced by the original charge plus a "mirror image" charge, equal in magnitude and opposite in sign, as far below the floor as the original charge is above it. The effect of this "image charge" is produced by the surface charge distribution induced on the floor by the presence of the original.

a. Find the electric field at (just above) the floor, as a function of horizontal distance p from the point on the floor directly under the original charge. Take the original charge to be a point +Q, a distance a above the floor. Ignore any effects of "walls" or "ceiling."

b. Find the surface charge density sigma(p) induced on the floor.

c. Calculate the total surface charge induced on the floor. HINT: Divide the floor into concentric rings, centered on the point directly under the original charge. Assume the floor extends "to infinity," i.e., very far.

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Solution:

Net field = 2ESinθ

Where E=electric field due to each charge

E=k*q/(a^2 + p^2)

Sinθ=a/(a^2 + p^2)^(1/2)

Total Electric field is given by:

E1 = k*q/(a^2 + p^2) *a/(a^2 + p^2)^(1/2)

=2Kqa/(a^2 + p^2)^(3/2)

As no field is there on the floor,

Therefore, field due to floor = E1

E1= σ / 2ε (σ = surface charge density)

σ =2ε (E1)

=2ε 2qaK/(a^2 + p^2)^(3/2)

= qa/Π(a^2 + p^2)^(3/2)

Now, to calculate total charge, divide the whole floor into concentric rings centered at point just below the charge.

Then the total

Q=∫σ ds

=∫ qa2 Π pdp/ Π(a^2 + p^2)^(3/2)

=2aq ∫ p dp/ (a^2 + p^2)^(3/2)

On integrating it we get,

Q=2q

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Thank you for passing by :)

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# Homework Help: Electrostatics Problem (Calculating Electric Field)

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