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(C) The electric field $E$ at a distance $z$ along the axis of a uniformly charged ring of radius $R$ is given by: $E = \frac{kQz}{(z^2 + R^2)^{3/2}}$

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$a$

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(C) The electric field $E$ at a distance $z$ along the axis of a uniformly charged ring of radius $R$ is given by: $E = \frac{kQz}{(z^2 + R^2)^{3/2}}$ To find the position where the electric field is maximum,we differentiate $E$ with respect to $z$ and set it to zero: $\frac{dE}{dz} = kQ \left[ \frac{(z^2 + R^2)^{3/2} - z \cdot \frac{3}{2}(z^2 + R^2)^{1/2} \cdot 2z}{(z^2 + R^2)^3} \right] = 0$ $(z^2 + R^2)^{3/2} - 3z^2(z^2 + R^2)^{1/2} = 0$ $(z^2 + R^2) - 3z^2 = 0$ $R^2 - 2z^2 = 0$ $z = \frac{R}{\sqrt{2}}$ Given the radius $R = a\sqrt{2}$,we substitute this into the expression: $z = \frac{a\sqrt{2}}{\sqrt{2}} = a$ Thus,the electric field is maximum at $z = a$.

Consider a circular loop that is uniformly charged and has a radius $R = a \sqrt{2}$. Find the position along the positive $z$-axis of the Cartesian coordinate system where the electric field is maximum,assuming the ring is placed in the $xy$-plane at the origin.

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