The number of electrons to be put on a spheri | vedclass

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

Question

(c) $E = \frac{1}{{4\pi {\varepsilon _0}}}.\frac{{ne}}{{{r^2}}}$ $==>$ $n = \frac{{E{r^2}}}{e}.4\pi {\varepsilon _0}$ $==> $ $n = \frac{{0.036 \

Vedclass · Accepted Answer

$2.5 	imes {10^5}$

Vedclass · Answer

(c) $E = \frac{1}{{4\pi {\varepsilon _0}}}.\frac{{ne}}{{{r^2}}}$ $==>$ $n = \frac{{E{r^2}}}{e}.4\pi {\varepsilon _0}$ $==> $ $n = \frac{{0.036 	imes 0.1 	imes 0.1}}{{9 	imes {{10}^9} 	imes 1.6 	imes {{10}^{ - 19}}}} = \frac{{360}}{{144}} 	imes {10^5}$= $2.5 	imes {10^5}$ $N/C$.

The number of electrons to be put on a spherical conductor of radius $0.1\,m$ to produce an electric field of $0.036\, N/C$ just above its surface is

Similar Questions

Two charges $q$ and $3 q$ are separated by a distance ' $r$ ' in air. At a distance $x$ from charge $q$, the resultant electric field is zero. The value of $x$ is :

Explain electric field and also electric field by point charge.

Whose result the whole electrostatic is ?

In the given figure electric field at center $O$ due to section $AB$ of uniformly charged ring is $\overrightarrow E$. What will be electric field at $O$ due to section $ACB$ ?

For given arrangement, where four charge fixed at ends of as quare as given, find value of additional charge $Q$ to be put on one of the vertices so that component of net electric field along the vertical symmetric axis is zero at every point on the vertical