Explore Westonci.ca, the leading Q&A site where experts provide accurate and helpful answers to all your questions. Join our Q&A platform and get accurate answers to all your questions from professionals across multiple disciplines. Discover in-depth answers to your questions from a wide network of professionals on our user-friendly Q&A platform.
Sagot :
[tex]E = \int\limits^2_{-3} {} \, \frac{27dy}{( 4 - y)^2}[/tex] is the integrals for the magnitude of the electric field.
A characteristic of electricity exists at every location in space when charge of any kind is present. The value of E, often known as the electric field strength, electric field intensity, or just the electric field, expresses the strength and direction of the electric field.
The formula for the electric field due to a point charge is:
[tex]E = \frac{kq}{r^2}[/tex]
Here q is the charge, r is the distance from the charge, and k = 8.99 × [tex]10^9 Nm^2/ C^2[/tex]
Imagine an infinitely - tiny portion of this charge distribution that has a charge [tex]dq[/tex]. This charge produces an electric field dE with magnitude:
[tex]\int\limits^ {} \, dE = \int\limits^ {} \, \frac{k dq}{r^2}[/tex]
[tex]E = \int\limits^ {} \, \frac{k dq}{r^2}[/tex]
Here, λ = 3.0 nC/m dq
= λ.dy dE = ∫k.dq
From y = -3 m to y = 2 m, there is a 3.0 nC/m uniform linear charge distributed along the y axis.
The integrals for the magnitude of the electric field at y = 4 m on the y axis:
[tex]E = k \lambda \int\limits^2_{-3} {} \, \frac{dy}{( 4 - y)^2}[/tex]
[tex]E = \int\limits^2_{-3} {} \, \frac{27dy}{( 4 - y)^2}[/tex]
Learn more about electric field here:
https://brainly.com/question/14372859
#SPJ1
We appreciate your visit. Our platform is always here to offer accurate and reliable answers. Return anytime. We hope this was helpful. Please come back whenever you need more information or answers to your queries. Thank you for choosing Westonci.ca as your information source. We look forward to your next visit.