What is the electric field at a distance x = | vedclass

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

Question

(C) The electric field $E$ at a distance $x$ on the axis of a charged ring is given by the formula:$E = \frac{KQx}{(R^2 + x^2)^{3/2}}$Given $x = \sqrt

Vedclass · Accepted Answer

$\frac{\sqrt{8} KQ}{27 R^2}$

Vedclass · Answer

(C) The electric field $E$ at a distance $x$ on the axis of a charged ring is given by the formula:$E = \frac{KQx}{(R^2 + x^2)^{3/2}}$Given $x = \sqrt{8} R$,we substitute this into the formula:$E = \frac{KQ(\sqrt{8} R)}{(R^2 + (\sqrt{8} R)^2)^{3/2}}$$E = \frac{\sqrt{8} KQR}{(R^2 + 8R^2)^{3/2}}$$E = \frac{\sqrt{8} KQR}{(9R^2)^{3/2}}$$E = \frac{\sqrt{8} KQR}{27 R^3}$$E = \frac{\sqrt{8} KQ}{27 R^2}$Since $\sqrt{8} = 2\sqrt{2}$,the expression can also be written as $\frac{2\sqrt{2} KQ}{27 R^2}$.

What is the electric field at a distance $x = \sqrt{8} R$ from the center on the axis of a charged ring with charge $Q$ and radius $R$?

Similar Questions

The energy density of an electric field is proportional to which of the following?

As shown in the figure,at what distance in $cm$ from point $A$ will the electric field be zero?

The diagram shows symmetrically placed rectangular insulators with uniformly charged distributions of equal magnitude. At the origin,the net electric field $\vec{E}_{net}$ is:

Two charged conducting spheres of radii $5 \ cm$ and $10 \ cm$ have equal surface charge densities. If the electric field on the surface of the smaller sphere is $E$,then the electric field on the surface of the larger sphere is

$ABC$ is an equilateral triangle. Charges $+q$ are placed at each corner. The electric intensity at $O$ will be

Vedclass Test Series

Exam Paper Generator

Online Exam Module