Get detailed and accurate responses to your questions on IDNLearn.com. Get the information you need from our community of experts, who provide detailed and trustworthy answers.
[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