Question 211: Electrostatics & Capacitance

Question

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

Accepted Answer

To determine the position along the positive $ z $-axis where the electric field due to a uniformly charged circular loop is at its maximum, we follow these steps:
Expression for Electric Field ($ E $):
The electric field $ E $ at a point along the $ z $-axis for a charged circular loop can be expressed as:
$    E = \frac{KQr}{(x^2 + R^2)^{3/2}}    $
Here, $ K $ is the Coulomb's constant, $ Q $ is the total charge, $ r $ is a constant involving charge distribution, $ x $ is the distance along the $ z $-axis, and $ R $ is the radius of the loop.
Maximizing the Electric Field:
To find where $ E $ is maximum, we take the derivative of $ E $ with respect to $ x $ and set it to zero:
$    \frac{dE}{dx} = 0    $
Solve for $ x $:
Solving the equation from the derivative, we find:
$    x = \frac{R}{\sqrt{2}}    $
Substitute Given Radius:
Given the radius $ R = a\sqrt{2} $, substituting into the expression for $ x $:
$    x = \frac{\sqrt{2}a}{\sqrt{2}} = a    $
Thus, the position along the positive $ z $-axis where the electric field is maximum is at $ x = a $.

Answer

a

Answer

$a / 2$

Answer

0

Answer

$$ a / \sqrt{2} $$

Electrostatics & Capacitance

Consider a circular loop that is uniformly charged and has a radius \mathrm{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 x y plane at the origin :

Choose the correct answer: