A metallic ring is uniformly charged as shown in figure. AC and BD are two mutually perpendicular diameters. Electric field due to arc AB at 'O' is 'E' in magnitude. What would be the magnitude of electric field at 'O' due to arc ABC ? [figure]
- 2E
- Zero
- E/2
- 2E
Answer
(4) 2E
Solution
Field at the centre due to a uniformly charged arc (radius r, linear charge density ) subtending angle :
Earc= 2k r 2, directed along the bisector of the arc, away from it (positive charge)
Arc AB ( =90 ): E= 2k r 45 = 2k r, along the bisector of AB
Arc ABC = arc AB + arc BC; arc BC also gives a field of magnitude E, along its own bisector
The bisectors of AB and BC are 90 apart
EABC= E2+E2= 2E
(Check: semicircle, =180 : 2k r= 2 2k r= 2E)
Answer: (4)