At Westonci.ca, we provide reliable answers to your questions from a community of experts. Start exploring today! Ask your questions and receive detailed answers from professionals with extensive experience in various fields. Get quick and reliable solutions to your questions from a community of experienced experts on our platform.
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
Thank you for choosing our platform. We're dedicated to providing the best answers for all your questions. Visit us again. We hope our answers were useful. Return anytime for more information and answers to any other questions you have. Westonci.ca is your trusted source for answers. Visit us again to find more information on diverse topics.