A 10-cm-long thin glass rod uniformly charged to 15.0 nC and a 10-cm-long thin p

Question

A 10-cm-long thin glass rod uniformly charged to 15.0 nC and a 10-cm-long thin plastic rod uniformly charged to - 15.0 nC are placed side by side, 3.80 cm apart. What are the electric field strengths E1 to Es 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 E1 Express your answer with the appropriate units 3.49x105 N ious Ans Correct Part B Specify the electric field strength E2 Express your answer with the appropriate units ValueUnits SubmitPre vious Answers Request Answer X Incorrect; Try Again; 5 attempts remaining Part C Specify the electric field strength E3 Express your answer with the appropriate units ValueUnits Submit Request Answer

Explanation / Answer

Given,

L = 10 cm ; Q = 15 nC ; L' = 10 cm ; Q' = 15 nC ; d = 3.8 cm ;

We know that, E = kQ/R^2

R = r sqrt (r^2 + (L/2)^2

E = 9 x 10^9 x 15 x 10^-9/(0.01 sqrt(0.01^2 + 0.05^2) = 2.65 x 10^5 N/C

r = 3.8 - 1 = 2.8 cm = 0.028 m

E = 9 x 10^9 x 15 x 10^-9/(0.028 sqrt(0.028^2 + 0.05^2) = 0.84 x 10^5

Enet = E + E' = (2.65 + 0.84) x 10^5 = 3.49 x 10^5 N/C

b)2cm

E = 9 x 10^9 x 15 x 10^-9/(0.02 sqrt(0.02^2 + 0.05^2) = 1.25 x 10^5 N/C

r = 3.8 - 2 = 1.8 cm = 0.018 m

E = 9 x 10^9 x 15 x 10^-9/(0.018 sqrt(0.018^2 + 0.05^2) = 1.41 x 10^5

Enet = E + E' = (1.25 + 1.41) x 10^5 = 2.66 x 10^5 N/C

Hence, E = 2.66 x 10^5 N/C

c)3 cm

E = 9 x 10^9 x 15 x 10^-9/(0.03 sqrt(0.03^2 + 0.05^2) = 0.77 x 10^5 N/C

r = 3.8 - 3 = 0.8 cm = 0.008 m

E = 9 x 10^9 x 15 x 10^-9/(0.008 sqrt(0.008^2 + 0.05^2) = 3.33 x 10^5

Enet = E + E' = (0.77 + 3.33) x 10^5 = 4.1 x 10^5 N/C

Hence, E = 4.1 x 10^5 N/C

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