An infinite non-conducting sheet has a surfac | vedclass

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

Question

(D) The electric field $E$ produced by an infinite non-conducting sheet is given by $E = \frac{\sigma}{2 \varepsilon_0}$.Given $\sigma = 7 	imes 10^{

Vedclass · Accepted Answer

$0.5$

Vedclass · Answer

(D) The electric field $E$ produced by an infinite non-conducting sheet is given by $E = \frac{\sigma}{2 \varepsilon_0}$.Given $\sigma = 7 	imes 10^{-7} 	ext{ C m}^{-2}$ and $\frac{1}{4 \pi \varepsilon_0} = 9 	imes 10^9 	ext{ SI units}$.We know that $\varepsilon_0 = \frac{1}{4 \pi 	imes 9 	imes 10^9} \approx 8.85 	imes 10^{-12} 	ext{ F m}^{-1}$.The relation between electric field and potential difference $\Delta V$ for a distance $\Delta r$ is $|E| = \frac{\Delta V}{\Delta r}$.Substituting the values:$\frac{\sigma}{2 \varepsilon_0} = \frac{\Delta V}{\Delta r}$$\Delta r = \frac{\Delta V 	imes 2 \varepsilon_0}{\sigma} = \frac{19.8 	imes 2 	imes 8.85 	imes 10^{-12}}{7 	imes 10^{-7}}$$\Delta r = \frac{19.8 	imes 17.7 	imes 10^{-12}}{7 	imes 10^{-7}} \approx 5 	imes 10^{-4} 	ext{ m} = 0.5 	ext{ mm}$.

Similar Questions

The figure represents two equipotential lines in the $x-y$ plane for an electric field. The $x$-component $E_{x}$ of the electric field in the space between these equipotential lines is, (in $V/m$)

An electric field $\vec{E} = (25 \hat{i} + 30 \hat{j}) \, NC^{-1}$ exists in a region of space. If the potential at the origin is taken to be zero,then the potential at $x = 2 \, m, y = 2 \, m$ is ...... $volt$.

Electric potential in a region is varying according to the relation $V = \frac{3x^2}{2} - \frac{y^2}{4}$,where $x$ and $y$ are in $m$ and $V$ is in $V$. Electric field intensity (in $N/C$) at a point $(1 \, m, 2 \, m)$ is:

In a certain region of space,the variation of potential with distance from the origin as we move along the $x$-axis is given by $V = 8x^2 + 2$,where $x$ is the $x$-coordinate of a point in space. The magnitude of the electric field at a point $(-4, 0)$ is .......... $V/m$.

The electric field $\vec E$ between two points is constant in both magnitude and direction. Consider a path of length $d$ at an angle $\theta = 60^o$ with respect to the field lines as shown in the figure. The potential difference between points $1$ and $2$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module