A 10-cm-long thin glass rod uniformly charged to 11.0nC and a 10-cm-long thin pl
ID: 2165749 • Letter: A
Question
A 10-cm-long thin glass rod uniformly charged to 11.0nC and a 10-cm-long thin plastic rod uniformly charged to - 11.0nC are placed side by side, 4.50 cm apart. What are the electric field strengths to at distances 1.0 cm, 2.0 cm, and 3.0 cm from the glass rod along the line connecting the midpoints of the two rods?(I.E. Find E1, E2, and E3)
Explanation / Answer
E=1/(4*pi*eo) * 2 ?/r is for infinite rod (L >>> y) --------------------------------- for bisector point (distance y from +rod and (0.04 - y) from negative rod) fields are E(y) = 9*10^9[2 ?/y] [(L/2) /sqrt{y^2+(L/2)^2}] direction (+x) E(0.04-y)=9*10^9[ -2?/(0.04-y)][(L/2) /sqrt{(0.04-y)^2+(L/2)^2}] note )- ve sign of lamda ? in 2nd equ >> indicates opposite direction of field due to -ve rod Net field at point (y) E(y-net) = E(y) (i) + E(0.04-y) ( - i) |E(y-net)} = |E(y)| - |E(0.04-y) --- (A) ======================= ? = Q/L = 10*10^-9/0.1 = C/meter ================= for y=1cm = 0.01 meter calculate E(0.01) then (0.04 - 0.01) by 2 equs and use (A) repeat for y=2 y=3
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.