Question

o -1 points Kat:PSE1 24 P 021 Often we have distributions of charge for which integrating to find the electric field may not be possible in practice. In such cases, we may be able to get a goed approximate solution by dividing the distribution into small but finite partices and taking the vector sum of the contributions of each. To see how this might work, consider a very thin rod of length 14 cm with uniform linear charge density λ-38.0 nC/m. Estimate the maritude of the electric field at a point P a atace d-70 cm hom the end of equal length as llustrated in the fiqure below for n- 4. Treat each segment as a particle whose distance from point P is measured from its center. Find estimates of Es for n 1,2, 4, and 8 segments. N/C N/C N/C N/C 1 2 34 Need Help? aad

Answer 1

The charge on a rod of length is given by -

Q = $\lambda$ L

where, $\lambda$ = linear charge density = 38 x 10^-9 C/m

L = length of a rod = 0.14 m

then, we get

Q = [(38 x 10^-9 C/m) (0.14 m)]

Q = 5.32 x 10^-9 C

Therefore, magnitude of an electric field due to charge, Q at point P located at a distance, r which will be given as -

E = k_e Q / r²

# For n = 1 segments, we have

E_P = k_e Q / [d + (L/2)]²

E_P = [(9 x 10⁹ Nm²/C²) (5.32 x 10^-9 C)] / {(0.07 m) + [(0.14 m) / 2]}²

E_P = [(47.8 Nm²/C) / (0.0196 m²)]

E_P = 2438.7 N/C

# For n = 2 segments, we have

E_P = k_e (Q/2) / [d + (3L/4)]² + k_e (Q/2) / [d + (L/4)]²

E_P = [(9 x 10⁹ Nm²/C²) (10.64 x 10^-9 C)] / {(0.07 m) + [(3 x 0.14 m) / 4]}² + [(9 x 10⁹ Nm²/C²) (10.64 x 10^-9 C)] / {(0.07 m) + [(0.14 m) / 4]}²

E_P = [(95.7 Nm²/C) / (0.030625 m²)] + [(95.7 Nm²/C) / (0.011025 m²)]

E_P = [(3124.9 N/C) + (8680.2 N/C)]

E_P = 11805.1 N/C