A thin rod, 50.1 cm long, is charged uniformly with a positive charge density of
ID: 1422814 • Letter: A
Question
A thin rod, 50.1 cm long, is charged uniformly with a positive charge density of 62.5 muC/m. The rod is placed along the y axis and is centered at the origin. A charge of +24.2 muC is placed 42.8 cm from the midpoint of the rod on the positive x axis. Calculate the electric field at the point on the.r axis that is halfway between the point charge and the centre of the rod. Answer: A rod of length 90.0 cm has a uniform linear charge density of 2.00 muC/m. Determine the electric field at the point P_1, located a distance ot 84.0 cm from the midpoint of the rod. Answer 15. Find the electric field at a point on the axis, P_2, a distance of 84.0 cm from one end of the rod. Answer: 16. What is the magnitude of the force per unit area between two infinite, uniformly charged plates with a surface charge density of +6.55 Times 10^-5 C/m^2 and -6.55 Times 10^-5 C/m^2), respectively, when the distance between the plates is 14.4 cm? Answer: 17. What is the force per unit area if the distance between the plates is doubled? Answer:Explanation / Answer
13. on x axis at distance x due to rod
E1 = kQ / x sqrt((L/2)^2 +x^2)
E1 = (9 x 10^9 x 62.5 x 10^-6 C/m x 0.501) / (0.214 sqrt(0.2505^2 + 0.214^2)) (i)
E1 = 4i x 10^6 N/C
due to point charge,
E2 = kQ / d^2
E2 = (9 x 10^9 x 24.2 x 10^-6 ) / (0.214^2) (-i)
= - 4.76i x 10^6 N/C
Enet = E1 + E2 = - 0.76 x 10^6 N/C .....Ans
magnitude = 0.76 x 10^6 N/C along -ve x axis.
14.
on the axis of perpendicular and passing through centre of rod at distance x due to rod
E1 = kQ / d sqrt((L/2)^2 +d^2)
E1 = (9 x 10^9 x 2 x 10^-6 C/m x 0.90) / (0.84 sqrt(0.84^2 + 0.45^2))
E1 = 20238.06 N/C
15. E = kQ / d(d + L)
E = (9 x 10^9 x 2 x 10^-6 x 0.90 ) / (0.84 (0.84 + 0.45))
E = 14950.17 N/C
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