Question

A disk of radius \(a\) has a total charge \(Q\) uniformly distributed over its surface. The disk has negligible thickness and lies in the \(x y\) plane. Throughout this problem, you may use the variable \(k\) in place of \(\frac{1}{4 \pi \epsilon_{0}}\)

Part A

What is the electric potential \(V(z)\) on the \(z\) axisas a function of \(z,\) for \(z>0\) ? Express your answer in terms of \(Q, z\), and \(a\). You may use \(k\) instead of \(\frac{1}{4 \pi \epsilon_{0}}\).

Part B

What is the magnitude \(E\) of the electric field on the \(z\) axis, as a function of \(z\), for \(z>0\) ? Express your answer in terms of someor all of the variables \(Q, z\), and \(a\). You may use \(k\) instead of \(\frac{1}{4 \pi \epsilon_{0}}\)

Answer 1

From the given figure,

Radius of the disk \(=a\)

Charge \(=\mathrm{Q}\)

Considering \(\sigma\) to be the area charge density, \(d q\) to be an infinitesimal charge element and da to be an infinitesimal radius element. then.

\(d q=\sigma^{*} 2^{*} \pi^{*} a^{*} d a->(1)\)

Part A answer:

Since, the forumal for electric potential is given by,

$$ V=\frac{k * Q}{r} $$

Considering infinitesimal version of above equation,

\(d V=\frac{k * d Q}{r}\)

Substituting (1) in the above equation,

\(V=k * \sigma * 2 * \pi * \int_{0}^{a} \frac{R^{\prime} * d R^{\prime}}{r}\)

where \(\mathrm{R}^{\prime}\) and \(\mathrm{dR}^{\prime}\) are the radius and infinitesimal radius of an infinitesimal uniformally distributed disk element

Integrating the above equation with the given limits,

\(V(z)=k * \sigma * 2 * \pi *\left(\sqrt{z^{2}+a^{2}}-z\right)\)

which is the required electric potential equation

Part B answer:

Since, the forumal for electric field is given by,

\(E=\frac{k * Q}{r^{2}}\)

Considering infinitesimal version of above equation,

\(d E=\frac{k * d Q}{r^{2}}\)

Substituting (1) in the above equation,

\(E=k * \sigma * 2 * \pi * z * \int_{0}^{a} \frac{R^{\prime} * d R^{\prime}}{\sqrt[3]{\left(z^{2}+R^{\prime 2}\right)}}\)

where R' and dR' are the radius and infinitesimal radius of an infinitesimal uniformally distributed disk element

Integrating the above equation with the given limits,

\(E(z)=k * \sigma * 2 * \pi *\left(1-\frac{z}{\sqrt{z^{2}+a^{2}}}\right)\)

which is the required electric field equation.