A thin half ring of radius 35 text{ cm} is un | vedclass

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

Question

(A) For a semi-circular ring of radius $R$ and total charge $Q$, the linear charge density is $\lambda = \frac{Q}{\pi R}$.The electric field $E$ at th

Vedclass · Accepted Answer

$2.14$

Vedclass · Answer

(A) For a semi-circular ring of radius $R$ and total charge $Q$, the linear charge density is $\lambda = \frac{Q}{\pi R}$.The electric field $E$ at the center is given by the formula $E = \frac{2k\lambda}{R}$, where $k = \frac{1}{4\pi\epsilon_0}$.Substituting $\lambda$, we get $E = \frac{2(1/4\pi\epsilon_0)(Q/\pi R)}{R} = \frac{Q}{2\pi^2 \epsilon_0 R^2}$.Given $E = 100 	ext{ V/m}$, $R = 0.35 	ext{ m}$, $\epsilon_0 = 8.85 	imes 10^{-12} 	ext{ C}^2/	ext{Nm}^2$, and $\pi = 3.14$.Rearranging for $Q$: $Q = E 	imes 2\pi^2 \epsilon_0 R^2$.$Q = 100 	imes 2 	imes (3.14)^2 	imes 8.85 	imes 10^{-12} 	imes (0.35)^2$.$Q = 200 	imes 9.8596 	imes 8.85 	imes 10^{-12} 	imes 0.1225$.$Q \approx 2.14 	imes 10^{-9} 	ext{ C} = 2.14 	ext{ nC}$.

Similar Questions

In the given figure, find the distance from point $A$ where the electric field is zero (in $cm$).

Two charges $Q$ are placed at two vertices of an equilateral triangle of side $a$. What is the electric field at the third vertex?

Two charges $Q$ and $4Q$ are separated by a distance of $6 \text{ cm}$. The distance of the point from $4Q$ at which the net electric field is zero is: (in $\text{ cm}$)

$A$ uniformly charged sphere has a total charge $Q$ and radius $R$. The electric field $E$ is a function of the distance $r$ from the center. Which of the following graphs represents this relationship?

Three identical point charges are placed at the vertices of an isosceles right-angled triangle as shown. Which of the numbered vectors coincides in direction with the electric field at the mid-point $M$ of the hypotenuse?

Vedclass Test Series

Exam Paper Generator

Online Exam Module