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Hint: Suppose two point charges are separated by a distance r. The two charges will attract each other. This force of attraction is called the electrostatic force of attraction. Use the formula for the magnitude of the electrostatic force between two charges and find the value of r.
Formula used:
$F=\dfrac{k{{q}_{1}}{{q}_{2}}}{{{r}^{2}}}$
Complete answer:
Suppose two point charges (+q and -q) are separated by a distance r. The two charges will attract each other. This force of attraction is called the electrostatic force of attraction. The magnitudes of the forces of attraction experienced by each of the charges are equal.
The magnitude of the electrostatic force between two charges is given as $F=\dfrac{k{{q}_{1}}{{q}_{2}}}{{{r}^{2}}}$ …. (i)
where k is a proportionality constant, ${{q}_{1}}$ and ${{q}_{2}}$ are the magnitudes of the two charges.
In this case, ${{q}_{1}}={{q}_{2}}=q$.
Substitute the values in equation (i).
$\Rightarrow F=\dfrac{kq\times q}{{{r}^{2}}}=\dfrac{k{{q}^{2}}}{{{r}^{2}}}$ …. (ii).
It is given that the magnitude of the force between the charges +q and –q is equal to 16N.
This means that F = 16N.
Substitute the value of F in equation (ii).
$\Rightarrow 16=\dfrac{k{{q}^{2}}}{{{r}^{2}}}$
$\Rightarrow {{r}^{2}}=\dfrac{k{{q}^{2}}}{16}$
$\Rightarrow r=\sqrt{\dfrac{k{{q}^{2}}}{16}}$.
(Since distance is always positive, we will not consider the negative value of the square root.)
$\Rightarrow r=\dfrac{q}{4}\sqrt{k}\text{ }metre$
Therefore, the distance between the two given charges is $\dfrac{q}{4}\sqrt{k}\text{ }metre$.
Hence, the correct option is B.
Note:
Note that the electrostatic force between two charges is not always attractive like the gravitational force between two masses.
The electrostatic force is attractive in nature when the two are unlike charges. That is one positive charge and the other negative charge.
The electrostatic force is repulsive in nature when the two are like charges. That is either both positive charges or both negative charges.
Formula used:
$F=\dfrac{k{{q}_{1}}{{q}_{2}}}{{{r}^{2}}}$
Complete answer:
Suppose two point charges (+q and -q) are separated by a distance r. The two charges will attract each other. This force of attraction is called the electrostatic force of attraction. The magnitudes of the forces of attraction experienced by each of the charges are equal.
The magnitude of the electrostatic force between two charges is given as $F=\dfrac{k{{q}_{1}}{{q}_{2}}}{{{r}^{2}}}$ …. (i)
where k is a proportionality constant, ${{q}_{1}}$ and ${{q}_{2}}$ are the magnitudes of the two charges.
In this case, ${{q}_{1}}={{q}_{2}}=q$.
Substitute the values in equation (i).
$\Rightarrow F=\dfrac{kq\times q}{{{r}^{2}}}=\dfrac{k{{q}^{2}}}{{{r}^{2}}}$ …. (ii).
It is given that the magnitude of the force between the charges +q and –q is equal to 16N.
This means that F = 16N.
Substitute the value of F in equation (ii).
$\Rightarrow 16=\dfrac{k{{q}^{2}}}{{{r}^{2}}}$
$\Rightarrow {{r}^{2}}=\dfrac{k{{q}^{2}}}{16}$
$\Rightarrow r=\sqrt{\dfrac{k{{q}^{2}}}{16}}$.
(Since distance is always positive, we will not consider the negative value of the square root.)
$\Rightarrow r=\dfrac{q}{4}\sqrt{k}\text{ }metre$
Therefore, the distance between the two given charges is $\dfrac{q}{4}\sqrt{k}\text{ }metre$.
Hence, the correct option is B.
Note:
Note that the electrostatic force between two charges is not always attractive like the gravitational force between two masses.
The electrostatic force is attractive in nature when the two are unlike charges. That is one positive charge and the other negative charge.
The electrostatic force is repulsive in nature when the two are like charges. That is either both positive charges or both negative charges.
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