Two point charges $+Q$ and $+q$ repel each other with a force of $100 \,N$. Keeping the distance between them unchanged, if $Q$ is increased by $10 \%$ and $q$ is decreased by $10 \%$, the force of repulsion between them will:

Two free positive charges $4q$ and $q$ are at a distance $l$ apart. What charge $Q$ is needed to achieve equilibrium for the entire system and where should it be placed from charge $q$?

Two particles of equal mass $m$ and charge $q$ are placed at a distance of $16 \, cm$. They do not experience any net force. The value of $\frac{q}{m}$ is

$A$ small uncharged conducting sphere is placed in contact with an identical sphere having a charge of $4 \times 10^{-8} \text{ C}$. They are then separated to a distance such that the force of repulsion between them is $9 \times 10^{-3} \text{ N}$. The distance between them is (Take $\frac{1}{4 \pi \varepsilon_0} = 9 \times 10^9 \text{ N m}^2 \text{ C}^{-2}$ in $SI$ units). (in $\text{ cm}$)

Three identical charges are placed on three vertices of a square. If the force acting between $q_1$ and $q_2$ is $F_{12}$ and between $q_1$ and $q_3$ is $F_{13}$,then $\frac{F_{13}}{F_{12}} = $ . . . . . . .

Three charges $q, Q$ and $+4q$ are placed in a straight line of length $d$ at points at distances $0, \frac{d}{2}$ and $d$ respectively. In order to make the net force on $q$ to be zero,the value of $Q$ should be