Q.

A thin non-conducting ring is placed in XOY plane with its axis along Z-axis and centre at origin. Ring is charged such that it acquires a charge density of λ=λ0cos⁡θCm⋅λ0 is constant and θ is angle made with positive direction of X-axis, measured anti-clockwise.Charge on the ring isX-component of electric field at centre of the ring isY-component of electric field at the centre of the ring is

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a

positive everywhere

b

negative everywhere

c

positive in 34th part and negative in 14th part

d

total charge on the ring is zero

e

kλ0R

f

2kλ0R

g

λ04ε0R

h

λ02ε0R

i

kλ0R

j

2kλ0R

k

λ04ε0R

l

0

answer is , , .

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Detailed Solution

Charge density λ=λ0cos⁡θ,cos⁡θ is positive when θ∈−π2,π2 and cos⁡θ is negative when θ∈π2,3π2So, charge on ring is as shownNow, for an elemental part subtending angle dθ at centre of ring field produced at centre will be dE, given bydE=kdqR2,dq=(Rdθ)λ=λ0cos⁡θ⋅RdθElectric field dE can be resolved into components dEx and dEy.dEx=dEcos⁡θdEy=dEsin⁡θand  Ex=∫02π dEx=λ04πε0R∫02π cos2⁡θdθ=λ04ε0RAlso, Ey=∫02π dEy=0⇒Enet = Net field at the centre of the ring=Ex2+Ey2=λ04ε0R⋅NCCharge density λ=λ0cos⁡θ,cos⁡θ is positive when θ∈−π2,π2 and cos⁡θ is negative when θ∈π2,3π2So, charge on ring is as shownNow, for an elemental part subtending angle dθ at centre of ring field produced at centre will be dE, given bydE=kdqR2,dq=(Rdθ)λ=λ0cos⁡θ⋅RdθElectric field dE can be resolved into components dEx and dEy.dEx=dEcos⁡θdEy=dEsin⁡θand  Ex=∫02π dEx=λ04πε0R∫02π cos2⁡θdθ=λ04ε0RAlso, Ey=∫02π dEy=0⇒Enet = Net field at the centre of the ring=Ex2+Ey2=λ04ε0R⋅NCCharge density λ=λ0cos⁡θ,cos⁡θ is positive when θ∈−π2,π2 and cos⁡θ is negative when θ∈π2,3π2So, charge on ring is as shownNow, for an elemental part subtending angle dθ at centre of ring field produced at centre will be dE, given bydE=kdqR2,dq=(Rdθ)λ=λ0cos⁡θ⋅RdθElectric field dE can be resolved into components dEx and dEy.dEx=dEcos⁡θdEy=dEsin⁡θand  Ex=∫02π dEx=λ04πε0R∫02π cos2⁡θdθ=λ04ε0RAlso, Ey=∫02π dEy=0⇒Enet = Net field at the centre of the ring=Ex2+Ey2=λ04ε0R⋅NC
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