Question

Two small objects X and Y are permanently separated by a distance 1 cm. Object X has a charge of $+1.0\mu C$ and object Y has a charge of $-1.0\mu C$. A certain number of electrons are removed from X and put onto Y to make the electrostatic force between the two objects an attractive force whose magnitude is 360 N. Number of electrons removed is :

• A. $8.4\times {10}^{13}$
• B. $6.25\times {10}^{12}$
• C. $4.2\times {10}^{11}$
• D. $3.5\times {10}^{10}$

Answer 1

Answer: (B)

$6.25\times {10}^{12}$

New attractive force is -360 N.
$-360N=\frac{K\left(q_1\right)\left(q_2\right)}{r^2}$
$K=9\times {10}^9$, $r={10}^{-2}m$
$-360N=\frac{K(1\times {10}^{-6}+ne)(-(1\times {10}^{-6})-ne)}{(1\times {10}^{-2}{)}^2}$
$\frac{360\times {10}^{-4}}{9\times {10}^9}=(1\times {10}^{-6}+ne{)}^2$
$\frac{4}{{10}^{12}}=(1\times {10}^{-6}+ne{)}^2$
$2\times {10}^{-6}=1\times {10}^{-6}+ne.$
$ne=1\times {10}^{-6}$
$n=1\times {10}^{-6}=\frac{1\times {10}^{-6}}{1\cdot 6\times {10}^{-19}}=6\cdot 25\times {10}^{12}$