A metallic ring is uniformly charged as shown in figure. AC and BD are two mutually perpendicular diameters. Electric field due to $\operatorname{arc} \mathrm{AB}$ to ' O ' is ' E ' is magnitude. What would be the magnitude of electric field at ' O ' due to $\operatorname{arc} \mathrm{ABC}$ ?

JEE Main · 2025 · 04 Apr 2025 · Shift 2 · Electrostatics

A metallic ring is uniformly charged as shown in figure. AC and BD are two mutually perpendicular diameters. Electric field due to

arc AB

to ' O ' is ' E ' is magnitude. What would be the magnitude of electric field at ' O ' due to

arc ABC

(A)2E
(B) $2 E$ ✓ Correct
(C) $E /2$
(D)Zero

Answer: (B)

Solution

Each quarter arc of the uniformly charged ring produces a field of magnitude

E

at the centre, directed along the bisector of that arc. The half ring

A B C

is made of the two adjacent quarter arcs

A B

and

B C

, whose field directions at the centre are mutually perpendicular. Adding these two equal and perpendicular contributions, the field at the centre due to arc

A B C

has magnitude

E^{2} + E^{2} = 2 E

JEE Main · 2025 · 04 Apr 2025 · Shift 2 · Single correct option

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