JEE physics

A uniformly charged ring of radius $R$ and total charge $Q$ lies in the $xy$-plane. Find the electric field at a point $P$ on the axis at distance $x$ from the center.

DDilaniJayawardene31 Asked 2mo ago 104 views 1 answer

A uniformly charged ring of radius $R$ and total charge $Q$ lies in the $xy$-plane. Find the electric field at a point $P$ on the axis at distance $x$ from the center.

1 Answer

VAVidaara Admin ✓ Vidaara Team ✓ Accepted · 2mo ago ▲ 23

By symmetry, only the axial component survives:

$$E = \frac{Qx}{4\pi\epsilon_0(R^2+x^2)^{3/2}}$$

Maximum at $x = \dfrac{R}{\sqrt{2}}$:

$$E = \frac{Qx}{4\pi\epsilon_0(R^2+x^2)^{3/2}}$$

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Discussion (4)

AK
Why do we take the positive value only in the last step?
Aditya Kumar · 2mo ago
R
Underrated solution. The way you set it up makes it almost obvious.
RiteshBasnet93 · 2mo ago
AR
How do we know the approximation is valid here?
Ananya Reddy · 2mo ago
VA
Good follow-up questions — remember to always state your assumptions in the JEE subjective section.
Vidaara Admin · Vidaara Team · 2mo ago
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