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Find the electric potential at the axis of a uniformly charged disc and use the potential to find the electric field at the same point.

Answer
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Hint: First consider a small part of that disc that has a radius dr. We know the equation for electric potential is V=14πε0qr, so using this equation for that small part of the disc we get the equation dV=14πε0dqr2+x2. Then integrate this equation and we get this equation V=σ2ε0[R2+x2x]. We know that the magnitude of the electric field is E=dVdx, then substitute the value of V in this equation to get this equation E=d(σ2ε0[R2+x2x])dx. Finally, solve this equation to reach the solution.

Complete answer:
Electric field – It is the space around a charge where the electrostatic force of attraction or repulsion can be felt by other charges and charged bodies.
Electric potential –Electric potential at any point is the work done in bringing a unit positive charge in an electric point to that point.
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Let’s consider a charged disc of a radius r with a surface charge density σ. Now let’s consider a point P from the center of the charged disc that is at a distance x from the center of the disc.
To calculate the electric potential at point P due to the charged disc, consider a small part of the disc which is of a radius dr.
So, the charge dq of the smaller disc of a radius dr is
dq= The surface charge density of the disc × Area of the smaller disc
dq=σ(2πrdr)
We know the electric potential at a point is given by
V=14πε0qr
Here, V= The electric potential at that point
ε0= The permittivity of free space
q= The magnitude of the charge
r= The distance between the charge and the point
So, the electric point at point P due to the charged disc of a radius dr is
dV=14πε0dqr2+x2
dV=14πε0σ(2πrdr)r2+x2
Integrating the equation, we get
0RdV=0R14πε0σ(2πrdr)r2+x2
0RdV=2σπ4πε00Rrdrr2+x2
V=2σπ4πε0[r2+x2]0R
V=2σπ4πε0[R2+x2(0)2+x2]
V=σ2ε0[R2+x2x]
We know that the magnitude of the electric field at any point is given by the following equation
E=dVdx
So, the magnitude of the electric field at point P due to a uniformly charged disc is
E=d(σ2ε0[R2+x2x])dx
E=σ2ε0[d[R2+x2]dxdxdx]
E=σ2ε0[2x2R2+x21]
E=σ2ε0[1xR2+x2]

Note:
By the equation for electric potential and the magnitude of the electric field, i.e. V=σ2ε0[R2+x2x] and E=σ2ε0[1xR2+x2] for the charged disc, that the electric potential is smallest near the disc and the magnitude of the electric field is maximum near the center.
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