Electric charges of +10,mu C,+5,mu C,-3,mu C, | vedclass

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

Question

(C) The potential $V$ at a point due to a charge $q$ at a distance $r$ is given by $V = \frac{kq}{r}$.For a square of side $a = \sqrt{2}\,m$,the dista

Vedclass · Accepted Answer

$1.8 	imes 10^5\,V$

Vedclass · Answer

(C) The potential $V$ at a point due to a charge $q$ at a distance $r$ is given by $V = \frac{kq}{r}$.For a square of side $a = \sqrt{2}\,m$,the distance $r$ from each corner to the centre is $r = \frac{a}{\sqrt{2}} = \frac{\sqrt{2}}{\sqrt{2}} = 1\,m$.The net potential at the centre is the algebraic sum of potentials due to individual charges: $V_{net} = V_1 + V_2 + V_3 + V_4 = \frac{k}{r} (q_1 + q_2 + q_3 + q_4)$.Substituting the values: $V_{net} = \frac{9 	imes 10^9}{1} 	imes (10 + 5 - 3 + 8) 	imes 10^{-6}$.$V_{net} = 9 	imes 10^9 	imes 20 	imes 10^{-6} = 180 	imes 10^3 = 1.8 	imes 10^5\,V$.

Electric charges of $+10\,\mu C$,$+5\,\mu C$,$-3\,\mu C$,and $+8\,\mu C$ are placed at the corners of a square of side $\sqrt{2}\,m$. The potential at the centre of the square is:

Similar Questions

Water rises to a height $h$ in a capillary tube at the surface of the Earth. On the surface of the Moon,the height of the water column in the same capillary tube will be:

$A$ uniform rod of mass $M$ and length $L$ is placed on a smooth horizontal surface. $A$ horizontal force $F$ is applied at one end of the rod perpendicular to its length. The acceleration of the centre of mass of the rod is:

When iron sulfide is heated in air,a sulfur oxide $A$ is formed. When $A$ is dissolved in water,it produces an acid. What is the basicity of this acid?

Arrange the following compounds in the increasing order of their dipole moment:
$(I)$ Toluene
$(II)$ $m$-Dichlorobenzene
$(III)$ $o$-Dichlorobenzene
$(IV)$ $p$-Dichlorobenzene

Suppose that the points $(h, k)$,$(1, 2)$,and $(-3, 4)$ lie on the line $L_1$. If a line $L_2$ passing through the points $(h, k)$ and $(4, 3)$ is perpendicular to $L_1$,then $\frac{k}{h}$ equals

Vedclass Test Series

Exam Paper Generator

Online Exam Module