A positive charge Q is placed on a conducting | vedclass

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

Question

(C) According to Gauss's law,the electric field $E$ at a distance $r$ from a point charge $q$ is given by $\oint E \cdot ds = \frac{q_{enclosed}}{\var

Vedclass · Accepted Answer

$\frac{q}{4 \pi \varepsilon_0 r^2}$

Vedclass · Answer

(C) According to Gauss's law,the electric field $E$ at a distance $r$ from a point charge $q$ is given by $\oint E \cdot ds = \frac{q_{enclosed}}{\varepsilon_0}$.For a Gaussian surface of radius $r$ (where $r Thus,$E(4 \pi r^2) = \frac{q}{\varepsilon_0}$.Solving for $E$,we get $E = \frac{q}{4 \pi \varepsilon_0 r^2}$.The charge $Q$ on the conducting shell does not contribute to the electric field inside the cavity because the field due to a spherically symmetric shell of charge is zero inside it.

$A$ positive charge $Q$ is placed on a conducting spherical shell with inner radius $R_1$ and outer radius $R_2$. $A$ particle with charge $q$ is placed at the center of the spherical cavity. The magnitude of the electric field at a point in the cavity,at a distance $r$ from the center,is

Similar Questions

Obtain an expression for the electric field at the surface of a charged conductor.

According to Gauss's Theorem,the electric field of an infinitely long straight wire is proportional to

Three infinitely long charge sheets are placed as shown in the figure. The electric field at point $P$ is

Two infinite sheets of uniform charge density $+\sigma$ and $-\sigma$ are parallel to each other as shown in the figure. The electric field is:

$A$ solid metallic sphere has a charge $+3Q$. Concentric with this sphere is a conducting spherical shell having charge $-Q$. The radius of the sphere is $a$ and that of the spherical shell is $b$ $(b > a)$. What is the electric field at a distance $R$ $(a < R < b)$ from the centre?

Vedclass Test Series

Exam Paper Generator

Online Exam Module