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Sagot :
(a) The magnitude of the electric field at point 5.5m is 2.35 x 10⁴ N/C.
(b) The magnitude of the electric field at point 2.5m is 5.18 x 10⁴ N/C.
Electric field at a point on the Gaussian surface
The magnitude of the electric field at a point on the cylindrical Gaussian surface is calculated as follows;
E = λ/2πε₀r
where;
- λ is linear charge density
- ε₀ is permitivity of free space
- r is the position of the charge
At a distance of 5.5 m
[tex]E = \frac{\lambda}{2\pi \varepsilon _0 r} \\\\E = \frac{7.2 \times 10^{-6}}{2\pi \times 8.85 \times 10^{-12} \times 5.5} \\\\E = 2.35 \times 10^4 \ N/C[/tex]
At a distance of 2.5 m
[tex]E = \frac{\lambda}{2\pi \varepsilon _0 r} \\\\E = \frac{7.2 \times 10^{-6}}{2\pi \times 8.85 \times 10^{-12} \times 2.5} \\\\E = 5.18 \times 10^4 \ N/C[/tex]
Thus, the magnitude of the electric field at points of 5.5m is 2.35 x 10⁴ N/C, and the magnitude of the electric field at points of 2.5m is 5.18 x 10⁴ N/C.
Learn more about electric field here: https://brainly.com/question/14372859
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