Point B is located along the axis of the ring and is equidis

Point B is located along the axis of the ring and is equidistant from the ring\'s center and the line. If a proton were released from rest at point A, it would remain at rest. Known values: D, R, Q, k (or epsilon_0), epsilon, m_electron Find an expression (containing only known values) for the ratio of the two linear charge densities, lambda_line/lambda_ring. Find an expression (containing only known values) for the resulting acceleration (both magnitude and direction) of an electron released from rest at point B.

Solution

electric field due to the line of uniformly distributed charge=(/(2₀x))

where x=3D

E_L=(_line/(2₀(3D)))=(_line/(6₀D))

electric field due to the thin ring of uniformly charged=[kQa/(x²+R²)^3/2]

_ring = Q/2R

E_R=[_ringR²/2₀(x²+R²)^3/2]=[_ringR²/2₀(D²+R²)^3/2]

E_L+E_R=0

(_line/(6₀D))+[_ringR²/2₀(D²+R²)^3/2]=0

_line/_ring=-[R²/2₀(D²+R²)^3/2]*(6₀D)

means _line is negative charged.

b)E_resultant=(_line/(2₀D))+[_ringR²/2₀(D²+R²)^3/2]=4[_ringR²/2₀(D²+R²)^3/2] =4[QR³/4₀(D²+R²)^3/2]

magnitude of acceleration=eE/m=e{4[QR³/4₀(D²+R²)^3/2] }/m_electron

net electric field is towards

it means it accelerate towards the line