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The force of repulsion between two point...

The force of repulsion between two point charges is F, when these are at a distance of `1 m`. Now the point charges are replaced by spheres of radii `25 cm` having the charge same as that of point charges. The distance between their centres is `1 m`, then compare the force of repulsion in two cases.

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To solve the problem, we need to compare the force of repulsion between two point charges and the force of repulsion between two charged spheres of the same charge and radius. ### Step-by-Step Solution: 1. **Understanding the Initial Condition**: - We start with two point charges \( q_1 \) and \( q_2 \) that are separated by a distance of \( r = 1 \, m \). - The force of repulsion between these point charges is given as \( F \). ...
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Knowledge Check

  • The force of repulsion between two point charges is F, when these are at distance 0.5m apart. Now the point charges are replaced by spheres of radii 5 cm each having the same charge as that of the respective point charge. The distance between their centres is again kept 0.5 m. Then the force of repulsion will

    A
    increase
    B
    decrease
    C
    remain F
    D
    become `(10 F)/9`
  • The force of repulsion between two point charges is F, when these are at a distance of 1 m apart Now the point charges are replaced by spheres of radii 25 cm having the same charge as that of point charge and same distance apart. Then the new force of repulsion will

    A
    Increase
    B
    decrease
    C
    Remain same
    D
    First increase then decrease
  • The force of repulsion between two point charges is F, when they are d distance apart. If the point charges are replaced by conducted sphereas each of radius r and the charges remain same. The separation between the centre of sphere isd, then force of repulsion between them is

    A
    Equal to F
    B
    Less than F
    C
    Greater than F
    D
    Cannot be sait
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