Electric Field Due To a Finite Line of Charge Points:2 A rod of length 50 cm has

Question

Electric Field Due To a Finite Line of Charge Points:2 A rod of length 50 cm has a uniform linear charge density of 6 HC/m. Determine the Electric Field at a point P located at a perpendicular distance 91 cm along a line of symmetry of the rod. 5.76E10 N/C Since the rod is not infinitely long, using Gauss's law does not make things simpler. Use superposition and integrate. (Your text, page 690, has a derivation of this setup.) Submit AnswerIncorrect. Tries 4/5 Previous Tries Find the electric field at P if P were located at a distance 91 cm from one end of the rod, as shown. 5.93E10 NC Submit AnswerIncorrect. Tries 1/5 Previous Tries

Explanation / Answer

charge on rod. Q = 6*10^-6*0.5= 3*10^-6 C

On the equitorial line, E = (k*Q/d)*(1/sqrt(d^2 + (L/2)^2))

= (9*10^9*3*10^-6/0.5)*(1/sqrt(0.5^2 + (0.91/2)^2) )

E= 79881.65 N/c

on the axis of line, E = (k*Q/L)*(1/d - 1/(d+L) )

= (9*10^9*3*10^-6/0.5)*(1/0.5 - 1/(0.5+0.91) )

= 69702 N/c

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