Two concentric rings, one of radius 'a' and the other of radius 'b' (b a) have the charge +q and - Physics - Electric Charges And Fields

Question

Two concentric rings, one of radius 'a' and the other of radius 'b'
(b>a) have the charge +q and (-2/5)-3/2q respectively
Find the ratio b/a if a charge particle is placed on the axis at
z=a is in equilibrium

Sanjay Singh · Accepted Answer

Formula of electric field at a point on the axis of the ring of radius R at a distance of x away from the centre is E=14π∈qx(x2+R2)32electric field due to ring of radius a at a distance a from centre  isE1=14π∈q×a(a2+a2)32electric field due to ring of radius b at a distance a from centre  isE2=14π∈(25)-32q×a(a2+b2)32Both the direction of E1 and E2 are in the opposite directions so for net electric field to be zero E1=E214π∈q×a(a2+a2)32=14π∈-(25)-32q×a(a2+b2)321(a2+a2)32=(25)-32(a2+b2)321(2a2)32=1(25)-32(a2+b2)32122a3=1(25)-32(a2+b2)3222a3=25×25(a2+b2)32squaring both sides8a6=4×2125(a2+b2)38×1258=(a2+b2a2)3125=(a2+b2a2)3raising power 13 on both sides5=a2+b2a2=1+b2a2b2a2=4ba=2RegardsDear
student

Two concentric rings, one of radius 'a' and the other of radius 'b' (b>a) have the charge +q and (-2/5)-3/2q respectively. Find the ratio b/a if a charge particle is placed on the axis at z=a is in equilibrium.

Two concentric rings, one of radius 'a' and the other of radius 'b' (b>a) have the charge +q and (-2/5)^-3/2q respectively. Find the ratio b/a if a charge particle is placed on the axis at z=a is in equilibrium.