For a uniformly charged ring of radius R,the | vedclass

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

Question

(B) The electric field $E$ at a distance $x$ from the center of a uniformly charged ring of radius $R$ is given by $E = \frac{k Q x}{(x^2 + R^2)^{3/2}

Vedclass · Accepted Answer

$R/\sqrt{2}$

Vedclass · Answer

(B) The electric field $E$ at a distance $x$ from the center of a uniformly charged ring of radius $R$ is given by $E = \frac{k Q x}{(x^2 + R^2)^{3/2}}$.To find the distance $h$ where the electric field is maximum,we set the derivative with respect to $x$ to zero: $\frac{dE}{dx} = 0$.Using the quotient rule: $\frac{d}{dx} \left[ \frac{x}{(x^2 + R^2)^{3/2}} \right] = \frac{(x^2 + R^2)^{3/2} - x \cdot \frac{3}{2}(x^2 + R^2)^{1/2} \cdot 2x}{(x^2 + R^2)^3} = 0$.This simplifies to $(x^2 + R^2) - 3x^2 = 0$,which gives $R^2 - 2x^2 = 0$.Therefore,$x^2 = R^2/2$,which implies $x = R/\sqrt{2}$.Thus,the maximum magnitude occurs at $h = R/\sqrt{2}$.

For a uniformly charged ring of radius $R$,the electric field on its axis has the largest magnitude at a distance $h$ from its centre. Then the value of $h$ is

Similar Questions

An electric field is given by $\vec{E} = e_1 \hat{i} + e_2 \hat{j} + e_3 \hat{k}$. $A$ charge $Q$ is moved by a displacement vector $\vec{r} = a \hat{i} + b \hat{j}$. The work done is:

Two point electric charges $+10^{-8} \text{ C}$ and $-10^{-8} \text{ C}$ are placed $0.1 \text{ m}$ apart. Find the magnitude of the total electric field at the centre of the line joining the two charges.

$A$ ring of radius $R$ is charged uniformly with a charge $+Q$. The electric field at a point on its axis at a distance $r$ from any point on the ring will be

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module