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
639k+ views
Hint: The two-point charges are of equal and opposite magnitude. The force between them is based on Coulomb's law of electrostatic force between the charges. At first, we will calculate the force between the charges by applying Coulomb's law and then for the transferred charges we will calculate the value of force by using the given condition.
Complete step by step answer:
Coulomb’s law of electrostatic force 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 along the line joining the two charges.
The formula is given as:
$F = k\dfrac{{{q_1}{q_2}}}{{{r^2}}}$
Here, $F$ is the force, $k$ is the proportionality constant, ${q_1},{q_2}$ are the charges,
$r$ is the distance between the charges.
We are given that the charges are $ + Q$ and $ - Q$ , therefore the force becomes
$F = k\dfrac{{Q( - Q)}}{{{r^2}}}$
$ \Rightarrow F = - k\dfrac{{{Q^2}}}{{{r^2}}}$
Taking magnitude only, we have
$ \Rightarrow F = k\dfrac{{{Q^2}}}{{{r^2}}}$ --equation $1$
After $25\% $ charge of A is transferred to B, the new charges are $Q - \dfrac{Q}{4}$ and $ - Q + \dfrac{Q}{4}$ respectively which is $\dfrac{{3Q}}{4}$ and $ - \dfrac{{3Q}}{4}$ .
Substituting these charges in the formula, we get
${F_0} = k\dfrac{{{{\left( {\dfrac{{3Q}}{4}} \right)}^2}}}{{{r^2}}}$
$ \Rightarrow {F_0} = k\dfrac{{9{{\left( Q \right)}^2}}}{{16{r^2}}}$ --equation $2$
From equation $1$ we know that
$F = k\dfrac{{{Q^2}}}{{{r^2}}}$
Substituting this value in equation $2$ we get
$\therefore{F_0} = \dfrac{{9F}}{{16}}$
Thus, option B is the correct option.
Note:k is the proportionality constant and the value of k is $$9 \times {10^9}\,N\,{m^2}\,{C^{ - 2}}$$ . k is known as Coulomb’s constant. The value of k totally depends on the medium in which the charged objects are placed. Please note, that it is easy to compare when the force is in terms of variables instead of substituting the value of Coulomb’s constant.
Complete step by step answer:
Coulomb’s law of electrostatic force 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 along the line joining the two charges.
The formula is given as:
$F = k\dfrac{{{q_1}{q_2}}}{{{r^2}}}$
Here, $F$ is the force, $k$ is the proportionality constant, ${q_1},{q_2}$ are the charges,
$r$ is the distance between the charges.
We are given that the charges are $ + Q$ and $ - Q$ , therefore the force becomes
$F = k\dfrac{{Q( - Q)}}{{{r^2}}}$
$ \Rightarrow F = - k\dfrac{{{Q^2}}}{{{r^2}}}$
Taking magnitude only, we have
$ \Rightarrow F = k\dfrac{{{Q^2}}}{{{r^2}}}$ --equation $1$
After $25\% $ charge of A is transferred to B, the new charges are $Q - \dfrac{Q}{4}$ and $ - Q + \dfrac{Q}{4}$ respectively which is $\dfrac{{3Q}}{4}$ and $ - \dfrac{{3Q}}{4}$ .
Substituting these charges in the formula, we get
${F_0} = k\dfrac{{{{\left( {\dfrac{{3Q}}{4}} \right)}^2}}}{{{r^2}}}$
$ \Rightarrow {F_0} = k\dfrac{{9{{\left( Q \right)}^2}}}{{16{r^2}}}$ --equation $2$
From equation $1$ we know that
$F = k\dfrac{{{Q^2}}}{{{r^2}}}$
Substituting this value in equation $2$ we get
$\therefore{F_0} = \dfrac{{9F}}{{16}}$
Thus, option B is the correct option.
Note:k is the proportionality constant and the value of k is $$9 \times {10^9}\,N\,{m^2}\,{C^{ - 2}}$$ . k is known as Coulomb’s constant. The value of k totally depends on the medium in which the charged objects are placed. Please note, that it is easy to compare when the force is in terms of variables instead of substituting the value of Coulomb’s constant.
Recently Updated Pages
What is BLO What is the full form of BLO class 8 social science CBSE

Explain the Treaty of Vienna of 1815 class 10 social science CBSE

10 examples of friction in our daily life

Draw a diagram of nephron and explain its structur class 11 biology CBSE

Write structures of the following compounds i 2 Chloro3methylpentane class 11 chemistry CBSE

A Paragraph on Pollution in about 100-150 Words

Trending doubts
Draw a labelled sketch of the human eye class 12 physics CBSE

Which are the Top 10 Largest Countries of the World?

Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE

Draw ray diagrams each showing i myopic eye and ii class 12 physics CBSE

Which is the correct genotypic ratio of mendel dihybrid class 12 biology CBSE

What is the Full Form of PVC, PET, HDPE, LDPE, PP and PS ?

