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

We have a uniformly charged ring of radius \( R=\sqrt{2} \) lying in the \(xy\)-plane (center at origin). We seek the point on the positive \(z\)-axis where the axial electric field is maximum.

Concept Used:

The magnitude of the electric field on the axis of a uniformly charged ring is

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

directed along \(+\hat z\) for \(z>0\). To maximize \(E(z)\) with respect to \(z\), we differentiate and set the derivative to zero.

Step-by-Step Solution:

Step 1: Maximize the function \( f(z)=\dfrac{z}{(R^2+z^2)^{3/2}} \) (the constant factor \( \dfrac{Q}{4\pi\varepsilon_0} \) does not affect 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: Solve for \(z\):

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

Final Computation & Result

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

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

The electric field is maximum at \( z=1 \) (along the positive \(z\)-axis).