A uniformly charged semicircular arc of radiu | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Consider a small element of length $dl = r d	heta$ at an angle $	heta$ with the vertical axis. The charge on this element is $dq = \lambda dl =

Vedclass · Accepted Answer

$\frac{\lambda}{2 \pi \varepsilon_0 r}$

Vedclass · Answer

(A) Consider a small element of length $dl = r d	heta$ at an angle $	heta$ with the vertical axis. The charge on this element is $dq = \lambda dl = \lambda r d	heta$.The electric field $dE$ at the centre due to this element is $dE = \frac{1}{4 \pi \varepsilon_0} \frac{dq}{r^2} = \frac{\lambda d	heta}{4 \pi \varepsilon_0 r}$.By symmetry,the horizontal components of the electric field cancel out. We only need to integrate the vertical component $dE \cos 	heta$ from $-\pi/2$ to $\pi/2$.$E = \int_{-\pi/2}^{\pi/2} \frac{\lambda \cos 	heta d	heta}{4 \pi \varepsilon_0 r} = \frac{\lambda}{4 \pi \varepsilon_0 r} [\sin 	heta]_{-\pi/2}^{\pi/2}$.$E = \frac{\lambda}{4 \pi \varepsilon_0 r} (1 - (-1)) = \frac{2 \lambda}{4 \pi \varepsilon_0 r} = \frac{\lambda}{2 \pi \varepsilon_0 r}$.

$A$ uniformly charged semicircular arc of radius $r$ has a linear charge density $\lambda$. What is the electric field at its centre? ($\varepsilon_0$ is the permittivity of free space)

Similar Questions

Two point charges $(+Q)$ and $(-2Q)$ are fixed on the $X$-axis at positions $x = a$ and $x = 2a$ from the origin, respectively. At what position on the axis is the resultant electric field zero?

The potential in an electric field varies as $V = (x^2 - y^2)$. The electric lines of force in the $X-Y$ plane are

Two point charges $-10 \mu C$ and $+5 \mu C$ are situated on the $X$-axis at $x=0$ and $x=\sqrt{2} \ m$. The point along the $X$-axis where the electric field becomes zero is:

The direction $(\theta)$ of the electric field $\vec{E}$ at point $P$ due to a uniformly charged finite rod is shown in the figure. Based on the geometry provided,determine the angle $\theta$.

Three charges $2q, -q, -q$ are located at the vertices of an equilateral triangle. At the centre of the triangle:

Vedclass Test Series

Exam Paper Generator

Online Exam Module