Show that the electric potential a distance z above the centre of a horizontal circular loop of radius R, which carries a uniform charge density per unit length λ.....

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Problem:

Show that the electric potential a distance z above the centre of a horizontal circular loop of radius R, which carries a uniform charge density per unit length λ, is given by

$V=\dfrac{\lambda R}{2\epsilon_0}\dfrac{1}{(z^2 + R^2)^{1/2}}$

Obtain an expression for the electrostatic field strength as a function of z.

Explanation:

The electric potential at $P(0, 0, z)$ is derived using the principle of superposition, summing contributions from all infinitesimal elements of the charged circular loop. The electrostatic field strength $E$ is then obtained as the negative gradient of the potential $V$ along the $z$ -axis.

For a uniform charge distribution on a circular loop:

Compute ( V ) using the distance formula $r = \sqrt{z^2 + R^2}$ .
Differentiate $V$ with respect to $z$ to get $E_z$ .

Solution:

Electric Potential $V$ :
The total charge on the loop is $Q = 2\pi R \lambda$ .
At a distance $z$ above the center of the loop, the electric potential is given by:
$V = \frac{1}{4\pi \epsilon_0} \int \frac{\lambda \, dl}{r}$

For a uniform loop, $dl = R \, d\theta$ , and $r = \sqrt{z^2 + R^2}$ (constant for all points on the loop).