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?

Arc AB is a quarter circle subtending

$$90^{\circ}$$

and its electric field at center is given as

E

Now consider arc ABC.

It consists of two quarter arcs:

• arc AB
• arc BC

Each subtends $$90^{\circ}$$, so each produces field of magnitude

E

at center.

For uniformly charged arc, field at center lies along bisector of the arc.

• For arc AB, field is along bisector of first quadrant diagonal toward southwest.
• For arc BC, field is along bisector of fourth quadrant diagonal toward northwest.

These two field vectors are perpendicular because the bisectors differ by

$$90^{\circ}$$

So resultant field due to arc ABC is

$$\sqrt{E^2+E^2}$$

$$=\sqrt{2}E$$