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Force of attraction between two point charges Q and –Q separated by d meter is. When these charges are placed on two identical spheres of radius ${\text{R = 0}}{\text{.3d}}$ whose entries are d meter apart, the force of attraction between them is ?
A. Greater than
B. Equal to
C. Less than
D. None of these
Answer
474.9k+ views
Hint:This question is based on the practical application of the ‘Coulomb’s law’ which quantifies the amount of electrostatic force of two stationary, electrically charged particles. This force can be attractive or repulsive in nature based on the charges. For positive charges or negative charges that are of the same sign it will be repulsive and for charges with different signs like positive and negative it will be attractive.
Complete step-by-step answer:
Initially, we have to understand the concept of Coulomb’s law which states that, “The force between two stationary charged objects is directly proportional to the product of the quantity of charges on the objects and inversely proportional to the square of the separation distance between the two objects.” And this force is called electrostatic or coulomb force.
\[{\text{F = k}}\dfrac{{{{\text{q}}_1}{{\text{q}}_2}}}{{{{\text{r}}^2}}}\] …………(1)
where, F is electric force; k is coulomb constant; \[{{\text{q}}_1},{\text{ }}{{\text{q}}_2}\] are charges and r is distance of separation.
As in the question given above, the charges and the distance of separation in both the systems is the same. So according to the equation (1), the force of attraction in both the systems should remain the same. Moreover, the electrostatic force between two bodies depends on their charges and distance of separation, not on their shape and configuration. Therefore, the force of attraction in both the cases remains the same.
(B) option is correct.
Note: Coulomb’s law is a universal law and follows the inverse square law. The electrostatic force is the strongest force in nature. It is a central force and may be attractive or repulsive depending upon the kind of charges but that has one exception that is when the difference between the magnitude of same charges (positive or negative) is very large then charge of same sign shows attractive nature.
Complete step-by-step answer:
Initially, we have to understand the concept of Coulomb’s law which states that, “The force between two stationary charged objects is directly proportional to the product of the quantity of charges on the objects and inversely proportional to the square of the separation distance between the two objects.” And this force is called electrostatic or coulomb force.
\[{\text{F = k}}\dfrac{{{{\text{q}}_1}{{\text{q}}_2}}}{{{{\text{r}}^2}}}\] …………(1)
where, F is electric force; k is coulomb constant; \[{{\text{q}}_1},{\text{ }}{{\text{q}}_2}\] are charges and r is distance of separation.
As in the question given above, the charges and the distance of separation in both the systems is the same. So according to the equation (1), the force of attraction in both the systems should remain the same. Moreover, the electrostatic force between two bodies depends on their charges and distance of separation, not on their shape and configuration. Therefore, the force of attraction in both the cases remains the same.
(B) option is correct.
Note: Coulomb’s law is a universal law and follows the inverse square law. The electrostatic force is the strongest force in nature. It is a central force and may be attractive or repulsive depending upon the kind of charges but that has one exception that is when the difference between the magnitude of same charges (positive or negative) is very large then charge of same sign shows attractive nature.
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