Point B is located along the axis of the ring and is equidistant from the ring\'

Question

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.

Explanation / Answer

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

where x=3D

EL=(line/(20(3D)))=(line/(60D))

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

ring = Q/2R

ER=[ringR2/20(x2+R2)3/2]=[ringR2/20(D2+R2)3/2]

EL+ER=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)Eresultant=(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] }/melectron

net electric field is towards

it means it accelerate towards the line

Related Questions

Navigate

• Browse (All)
• Browse P
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.