Question

A point charge q0 moves from point A to B in a semicircular path of diameter 2a, then to point C in a semicircular path of diameter a, and finally to point D in a circular path of radius a/2, as shown. Point A is at a distance 6a apart from infinitely long wire having linear charge density 2. All circular paths and wires are in the same plane. Minimum work done to move the charge from A to D is: + + B o ***** IC P A 60° 01. a/2 ko a 2a + ба 902 24 (a) -log 6 202 (b) 21TE 1052 19 2TTE 902 24 log 2πες 12 (d) None of these.

Answer 1

I is linear charge ty density of straight D wire. P tttttttttt A PA= 6a 600 BC=a AB= 20 OCE IN O, is foot of perpendicular from D to PA.

L DOO, = 60° 00,= OD X cos <DOO, a 4 → 09=]x cos 60° = x $100,= .:.00= 06-00, - 4-4 -9C= 4 :: P0. = PC -0,6 = (PA-(CA)) – 0,6 a po, = 160-(AB-BC))- >> POI = Ga-(20-2) - 4 =6a-a-q 4 4 PO = iga 4 confermately Sence, electric field due to long stepmaight caire is radially outward So electric potential mill need change by moving from D to Oy, because Do, is perpendicular to the electric field. . Potential at O, and D is same.

a. No=vo, Let patential at A is VA Electric field at a distance from the wire is E=4 216 1809 moy dr 2nfor i for small displacement dy towards the wire, dr change in patential is du = _ .dr =- Ellafl cos1880 » dv=-E(-dr)x(-1) caires decreasong soldi I =-ov) s dv= Ear » du=- Jau = a) it on Lam) - (No, -VA) = -1 [enl >> Vor-VA = where, A= PA=6a and % = Po, = 190 gro V [nr] ro g A [enporn in anto ror

No, VA V un (64x4 A 2nGo 194 No, VA D 26 in 24 19 ... menemum work done by external agent in rom A to D is Wege =AU = U6-UA = % Vp40 VA = 46( V7-VA) moving a charge fro म A

(^. Voi > West (No, VA) Vo, = VD) sollen eset 24 19 anto So, correct option is @