(C) In the first case,by Coulomb's law,the force between the spheres is:$F_1 = \frac{k Q_1 Q_2}{d^2}$When the two identical spheres are made to touch,

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(C) In the first case,by Coulomb's law,the force between the spheres is:$F_1 = \frac{k Q_1 Q_2}{d^2}$When the two identical spheres are made to touch,the total charge is redistributed equally between them. The new charge on each sphere is:$q = \frac{Q_1 + Q_2}{2}$Since $Q_1 >> Q_2$,we can approximate $q \approx \frac{Q_1}{2}$.After separating them to the original distance $d$,the new force $F_2$ is:$F_2 = \frac{k q^2}{d^2} = \frac{k (Q_1/2)^2}{d^2} = \frac{k Q_1^2}{4d^2}$Now,calculating the ratio $F_1/F_2$:$\frac{F_1}{F_2} = \frac{(k Q_1 Q_2 / d^2)}{(k Q_1^2 / 4d^2)} = \frac{4 Q_1 Q_2}{Q_1^2} = \frac{4 Q_2}{Q_1}$

Class 12 Physics Electric Charges and Fields Mix Examples-Electric Charges and Fields

Difficult

Two identical small spheres carry charges of $Q_1$ and $Q_2$ with $Q_1 >> Q_2$. The charges are $d$ distance apart. The force they exert on one another is $F_1$. The spheres are made to touch one another and then separated to distance $d$ apart. The force they exert on one another now is $F_2$. Then $F_1/F_2$ is

A
$\frac{4Q_1}{Q_2}$
B
$\frac{Q_1}{4Q_2}$
C
$\frac{4Q_2}{Q_1}$
D
$\frac{Q_2}{4Q_1}$

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Mix Examples-Electric Charges and Fields

$A$ long cylindrical shell carries positive surface charge $\sigma$ in the upper half and negative surface charge $-\sigma$ in the lower half. The electric field lines around the cylinder will look like the figure given in:

MediumJEE MAIN 2015

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Two particles $X$ and $Y$, of equal mass and with unequal positive charges, are free to move and are initially far away from each other. With $Y$ at rest, $X$ begins to move towards it with initial velocity $u$. After a long time, finally

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Two identical conducting spheres $A$ and $B$ carry equal charge. They are separated by a distance much larger than their diameter,and the force between them is $F$. $A$ third identical conducting sphere,$C$,is uncharged. Sphere $C$ is first touched to $A$,then to $B$,and then removed. As a result,the force between $A$ and $B$ would be equal to

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Two free point charges $+q$ and $+4q$ are a distance $R$ apart. $A$ third charge is so placed that the entire system is in equilibrium. Then the third charge is :-

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Two identical charged spheres suspended from a common point by two massless strings of lengths $l$ are initially at a distance $d$ $(d << l)$ apart because of their mutual repulsion. The charges begin to leak from both the spheres at a constant rate. As a result,the spheres approach each other with a velocity $v$. Then $v$ varies as a function of the distance $x$ between the spheres,as:

DifficultNEET 2016

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$A$ long cylindrical shell carries positive surface charge $\sigma$ in the upper half and negative surface charge $-\sigma$ in the lower half. The electric field lines around the cylinder will look like the figure given in:

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