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Question

# Find the electric field on the axis of a uniformly charged disc.

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Solution

## Consider a disc of uniform surface charge density σ.Let us calculate the electric field due to a charged ring of radius r, from the centre and having a width, dr. Here we are using the expression for electric field intensity for a charged ring of radius r at a point on the axis at a distance x from the centre,E=kxq(x2+r2)3/2 directed along the axis outwards from the centre.Thus electric field due to elementary ring,dE=kxdq(x2+r2)3/2, directed along the line OP.Now, the area of the ring, dS=2πr.dr.Charge on the elementary ring⟹dq=σ2πede,Thus, electric field due to entire disc|E|0=∫dE=kxR∫0σ2πrdr(x2+r2)3/2=14π∈0x.σπR∫02r(x2+r2)3/2drE=σ2∈0[1−x(x2+R2)1/2]=σ1ε0(1−cosθ), where θ= semi vertical angle subtend by the disc at R.Also, as R⟶∞,E=σ2∈0, which is the electric field in front of an infinite plane sheet of charge,

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