Phys 134: Introduction to Electricity and Magnetism Homeworkog Problem 23.9 Prob

Question

Phys 134: Introduction to Electricity and Magnetism Homeworkog Problem 23.9 Problem 23.9 Part A A 10-cm-long thin glass rod uniformly charged to 15.0 nCand a Speciy the electric field strengthEh 10 cm-ong thin plastic rod uniformly charged to are placed side by side, 430 cm apart. What are the electric field strengths Eh to Express your answer with the appropriate units. E, at distances 10 am, 20 cm and 3.0 from the glass rod along the Ine connecting the midpoints ofthe two rods? Value Units My Answers CheUR Part B Specify the electric field strength Eh Express your answer with the appropriate units. Value Units My Answers GuveUg Part C Specfy the electric feld strength E, Express your answer with the appropriate units. Value Units 75233696 https llsession assignment Problem

Explanation / Answer

The electric filed due to point charge is

dE = k dq/ r^2

since dq = lamda dy

The x- component

dEx = dE cos theta = dE ( x/r) = k x dq/ r^3

r = sqrt x^2 y^2

dEx =  k x dq/ r^3= k x * lamda dy/ ( x^2 + y^2 ) ^3/2

integrate with boundaries - l/2 to + l/2

Ex = k x lamda -l/2 integral to l/2 dy/ ( x^2 + y^2 ) ^3/2

since lamda Q/l

Ex = k x lamda ( y/ x^2 ( x^2 + y^2)^1/2) l/2 to -l/2

Ex = k Q/ x sqrt x^2 + l/2)^2

for 1 cm

E1 = 9 * 10^9 ( 15 * 10^-9)/0.01 sqrt( 0.01)^2 + (0.05)^2 = 2.64 * 10^5 N/C ( left)

=  9 * 10^9 ( 15 * 10^-9)/0.033 sqrt( 0.033)^2 + (0.05)^2 =0.68 * 10^5N/C ( right)

E_total = left- right =1.96* 10^5 N/C

at point 2.0cm

E2 = 9 * 10^9 ( 15 * 10^-9)/0.02 sqrt( 0.02)^2 + (0.05)^2 = 1.25* 10^5 N/C ( left)

=  9 * 10^9 ( 15 * 10^-9)/0.023sqrt( 0.023)^2 + (0.05)^2 =1.06 * 10^4 N/C ( right)

E_total = left- right =0.19* 10^5 N/C

at point 3.0cm

E3 = 9 * 10^9 ( 15 * 10^-9)/0.03 sqrt( 0.03)^2 + (0.05)^2 = 7.7174 * 10^4 N/C ( left)

=  9 * 10^9 ( 15 * 10^-9)/0.013 sqrt( 0.013)^2 + (0.05)^2 =20.10 * 10^4 N/C ( right)

E_total = left- right =-12.38* 10^4 N/C

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