Question

Two-point charges A and B, having charges +Q and -Q respectively, are placed at certain distance apart and force acting between them is F. If 25% charge of A is transferred to B, then force between the charges becomes
A. F
B. $\dfrac { 9F }{ 16 }$
C. $\dfrac { 16F }{ 9 }$
D. $\dfrac { 4F }{ 3 }$

Answer 1

Hint: As two-point charges of equal and opposite magnitude are present, it is based on Coulomb's law of electrostatic force between the charges. At first, the force between the charges is calculated by applying Coulomb's law and then for the transferred charges.The calculated values of force are compared to find the answer.

Formula used: $F=k\dfrac { { q }_{ 1 }{ q }_{ 2 } }{ { r }^{ 2 } }$

Complete step by step answer:
The formula is given by Coulomb’s law of electrostatic force which states that the force of attraction or repulsion between the point charges is directly proportional to the product of charges and inversely proportional to the square of the distance between them. The direction of force is always along the line that joins the two charges.
The given charges are +Q and -Q. So, q1 and q2 in the equation can be commonly taken as Q.
$F=k\dfrac { { Q }^{ 2 } }{ { r }^{ 2 } }$
Where, k is the proportionality constant and
     r is the assumed distance between the charges.
If 25% of charge from A is transferred to point B, there is a reduction of $\dfrac { Q }{ 4 }$ from charge of A and addition of $\dfrac { Q }{ 4 }$ to the charge of B.
${ q }_{ A }=Q-\dfrac { Q }{ 4 }$ and ${ q }_{ B }=-Q+\dfrac { Q }{ 4 }$
Which gives, ${ q }_{ A }=\dfrac { 3Q }{ 4 }$ and ${ q }_{ B }=-\dfrac { 3Q }{ 4 }$
Now substituting qA and qB in the formula,
${{F}_{1}}=k\dfrac{{{q}_{A}}{{q}_{B}}}{{{r}^{2}}}$
${{F}_{1}}=k\dfrac{{{\left( \dfrac{3Q}{4} \right)}^{2}}}{{{r}^{2}}}$
On squaring and solving, we get
${{F}_{1}}=\dfrac{9}{16}\dfrac{k{{Q}^{2}}}{{{r}^{2}}}$
${ F }_{ 1 }=\dfrac { 9 }{ 16 } F$ (since, $F=k\dfrac { { Q }^{ 2 } }{ { r }^{ 2 } }$ )
Therefore, the correct answer for the given question is option (B).

Note: k is the proportionality constant with value of about $\dfrac{1}{4\pi {{\in }_{0}}}$=$9\times {{10}^{9}}N{{m}^{2}}{{c}^{-2}}$ . On substituting q1 andq2 as 1C and r =1m in the equation. We get,
$F=\left( 9\times {{10}^{9}} \right)\dfrac{1}{{{1}^{2}}}=9\times {{10}^{9}}N$