A solid uncharged conducting sphere has a rad | vedclass

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

Question

(B) $1$. The conducting sphere is uncharged,meaning the net charge on the sphere is $0$.$2$. When a point charge $+Q$ is placed inside the hollow cavi

Vedclass · Accepted Answer

$\frac{kQ}{16a^2}$

Vedclass · Answer

(B) $1$. The conducting sphere is uncharged,meaning the net charge on the sphere is $0$.$2$. When a point charge $+Q$ is placed inside the hollow cavity,an induced charge of $-Q$ appears on the inner surface of the cavity to maintain electrostatic equilibrium.$3$. Since the sphere is uncharged,a charge of $+Q$ must appear on the outer surface of the sphere to ensure the total charge remains $0$.$4$. For any point outside the sphere (like point $P$ at $r = 4a$),the electric field is determined by the total charge enclosed by a Gaussian surface passing through $P$.$5$. The total charge enclosed by a spherical Gaussian surface of radius $4a$ is the sum of the point charge $+Q$ inside the cavity and the induced charge $+Q$ on the outer surface of the conductor.$6$. Total charge $Q_{net} = (+Q) + (-Q) + (+Q) = +Q$.$7$. By Gauss's Law,the electric field at a distance $r = 4a$ from the center is $E = \frac{k Q_{net}}{r^2} = \frac{k Q}{(4a)^2} = \frac{kQ}{16a^2}$.

Similar Questions

$A$ square Gaussian surface is placed in the $y-z$ plane. Its axis is along the $x-$axis and its center is at the origin. Two identical charges,each $Q$,are placed at points $(a, 0, 0)$ and $(-a, 0, 0)$. If each side length of the square is $2a$,then the electric flux passing through the square is:

$A$ metallic sphere is kept in between two oppositely charged plates. The most appropriate representation of the field lines is

The net electric flux due to a uniform electric field of $3 \times 10^3 \hat{i} \text{ NC}^{-1}$ through a cube of side $20 \text{ cm}$ oriented such that its faces are parallel to the coordinate planes is

Consider the charge configuration and spherical Gaussian surface as shown in the figure. When calculating the flux of the electric field over the spherical surface,the electric field will be due to:

Vedclass Test Series

Exam Paper Generator

Online Exam Module