Question

Consider a circular loop that is uniformly charged and has a radius $a\sqrt{2}$. 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 xy plane at the origin:

• A. a/2
• B. $\frac{a}{\sqrt{2}}$
• C. a
• D. 0

Answer 1

Answer: (C)

a

Answer 2

The electric field on the z-axis for a uniformly charged ring of radius $R$ is given by

$$E(z) = k \frac{Q\, z}{\left(z^2 + R^2\right)^{3/2}}.$$

Here, $R = a\sqrt{2}$.

To find the maximum, set

$$f(z) = \frac{z}{(z^2+R^2)^{3/2}},$$

and differentiate with respect to $z$:

$$\frac{df}{dz} = \frac{(z^2+R^2)^{3/2} - 3z^2(z^2+R^2)^{1/2}}{(z^2+R^2)^3} = \frac{(z^2+R^2) - 3z^2}{(z^2+R^2)^{5/2}}.$$

Setting the numerator equal to zero:

$$(z^2+R^2) - 3z^2 = 0 \quad \Rightarrow \quad R^2 - 2z^2 = 0 \quad \Rightarrow \quad z^2 = \frac{R^2}{2}.$$

Substitute $R^2 = 2a^2$:

$$z^2 = \frac{2a^2}{2} = a^2 \quad \Rightarrow \quad z = a \quad (\text{since } z > 0).$$

Thus, the maximum electric field occurs at $z = a$.