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JEE MAIN 2025
02-04-2025 SHIFT-2
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 :
Select the correct option:
A
$a / \sqrt{2}$
B
a
C
$0$
D
$a / 2$
✓ Correct! Well done.
✗ Incorrect. Try again or view the solution.
Solution
$E=\frac{K Q r}{\left(x^{2}+R^{2}\right)^{3 / 2}}$
$\frac{d E}{d x}=0$
$\therefore x=\frac{R}{\sqrt{2}}=\frac{\sqrt{2} a}{\sqrt{2}}=a$
Question Tags
JEE Main
Physics
Easy
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