Two point charges +9e and +e are placed at a | vedclass

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

Question

(A) Let the charge $q$ be placed at a distance $x$ from the charge $+9e$ along the line joining the two charges.For the charge $q$ to be in equilibriu

Vedclass · Accepted Answer

$12\, cm$ from $+9e$

Vedclass · Answer

(A) Let the charge $q$ be placed at a distance $x$ from the charge $+9e$ along the line joining the two charges.For the charge $q$ to be in equilibrium,the electrostatic force exerted by $+9e$ must be equal in magnitude and opposite in direction to the force exerted by $+e$.Using Coulomb's law,$F = \frac{K q_1 q_2}{r^2}$,we set the forces equal:$\frac{K(9e)(q)}{x^2} = \frac{K(e)(q)}{(16-x)^2}$Canceling $K, e,$ and $q$ from both sides:$\frac{9}{x^2} = \frac{1}{(16-x)^2}$Taking the square root on both sides:$\frac{3}{x} = \frac{1}{16-x}$Solving for $x$:$3(16-x) = x$$48 - 3x = x$$4x = 48$$x = 12\, cm$Thus,the charge $q$ must be placed at a distance of $12\, cm$ from the $+9e$ charge.

Column-$I$ (Molecules)	Column-$II$ (Number of lone pairs on central atom)
$A$. $NH_3$	$1$. Two
$B$. $H_2O$	$2$. Three
$C$. $XeF_2$	$3$. Zero
$D$. $CH_4$	$4$. Four
	$5$. One

Two point charges $+9e$ and $+e$ are placed at a distance of $16\, cm$ apart. Where should a third charge $q$ be placed between them so that it remains in equilibrium?

Similar Questions

Which of the following plants is commonly used as green manure?

Let $A(3-i)$ and $B(2+i)$ be two points in the Argand plane. If the point $P$ represents the complex number $z=x+iy$,which satisfies $|z-3+i|=|z-2-i|$,then the locus of the point $P$ is

Match the following molecules in Column-$I$ with the number of lone pairs on their central atom in Column-$II$.
Column-$I$ (Molecules) Column-$II$ (Number of lone pairs on central atom)
$A$. $NH_3$ $1$. Two
$B$. $H_2O$ $2$. Three
$C$. $XeF_2$ $3$. Zero
$D$. $CH_4$ $4$. Four
$5$. One

If $a = \hat{i} + \hat{j} + t \hat{k}$ and $b = \hat{i} + 2 \hat{j} + 3 \hat{k}$,then the values of $t$ for which $(a+b)$ and $(a-b)$ are perpendicular are:

$CrO_4^{2-}$ (Yellow) changes to $Cr_2O_7^{2-}$ (orange) in $pH = x$ and vice versa in $pH = y$,hence $x$ and $y$ are respectively:

Vedclass Test Series

Exam Paper Generator

Online Exam Module