Question

Consider a circular loop that is uniformly charged and has a radius $ \sqrt{2} $. Find the position along the positive $ z $-axis of the cartesian coordinate system where the electric field is maximum if the ring was assumed to be placed in the $ xy $-plane at the origin:

Hint:

For maximum electric field along the axis of a charged circular loop, set the derivative of the electric field with respect to \( x \) to zero.

Answer 1

Correct Answer: 1

A uniformly charged ring of radius \( R=\sqrt{2} \) is centered at the origin in the \(xy\)-plane. We aim to find the point on the positive \(z\)-axis where the axial electric field is maximized.

Concept Used:

The electric field magnitude along the axis of a uniformly charged ring is given by

\[ E(z)=\frac{1}{4\pi\varepsilon_0}\,\frac{Q\,z}{\big(R^2+z^2\big)^{3/2}}, \]

and it is directed along \(+\hat z\) for \(z>0\). To find the maximum of \(E(z)\), we compute its derivative with respect to \(z\) and set it to zero.

Step-by-Step Solution:

Step 1: We maximize the function \( f(z)=\dfrac{z}{(R^2+z^2)^{3/2}} \), as the constant factor \( \dfrac{Q}{4\pi\varepsilon_0} \) does not influence the location of the maximum.

\[ \frac{d}{dz}\big[\ln f(z)\big]=\frac{d}{dz}\left(\ln z-\frac{3}{2}\ln(R^2+z^2)\right)=0. \] \[ \Rightarrow \frac{1}{z}-\frac{3}{2}\cdot\frac{2z}{R^2+z^2}=0 \;\;\Longrightarrow\;\; \frac{1}{z}-\frac{3z}{R^2+z^2}=0. \]

Step 2: We solve the equation for \(z\):

\[ (R^2+z^2)-3z^2=0 \;\;\Longrightarrow\;\; R^2-2z^2=0 \;\;\Longrightarrow\;\; z^2=\frac{R^2}{2}. \] \[ \Rightarrow z=\frac{R}{\sqrt{2}}\quad(\text{for } z>0). \]

Final Computation & Result

Given \(R=\sqrt{2}\),

\[ z=\frac{\sqrt{2}}{\sqrt{2}}=1. \]

The electric field attains its maximum value at \( z=1 \) on the positive \(z\)-axis.