A 10-cm-long thin glass rod uniformly charged to 11.0 nC and a 10-cm-long thin p
ID: 1593225 • Letter: A
Question
A 10-cm-long thin glass rod uniformly charged to 11.0 nC and a 10-cm-long thin plastic rod uniformly charged to -11.0 nC are placed side by side, 3.60 cm apart. What are the electric field strengths E^1 to E^3 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? Part A Specify the electric field strength E^1 Express your answer with the appropriate units. Part B Specify the electric field strength E^2 Express your answer with the appropriate units.Explanation / Answer
E=1/(4*pi*eo) * 2 /r is for infinite rod (L >>> y)
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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)
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= Q/L = 10*10^-9/0.1 = C/meter
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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
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