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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)
    9F/16
  • c)
    16F/9
  • d)
    4F/3
Correct answer is option 'B'. Can you explain this answer?
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Two point charges A and B, having charges +Q and –Q respectively...


If 25% of charge of A transferred to B then

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Two point charges A and B, having charges +Q and –Q respectively...
-Q respectively, are placed at a distance d apart from each other. The electric field at a point P on the line joining the two charges depends on the distance of the point P from each of the charges.

Let the distance of point P from charge A be x, and the distance from charge B be (d-x). By Coulomb's law, the electric field due to charge A at point P is given by:

Ea = (1/(4πε0)) * (Q/x^2)

where ε0 is the permittivity of free space. Similarly, the electric field due to charge B at point P is given by:

Eb = (1/(4πε0)) * (-Q/(d-x)^2)

The negative sign in front of Q indicates that the electric field due to charge B is in the opposite direction to that due to charge A.

The total electric field at point P is the vector sum of the electric fields due to the two charges:

E = Ea + Eb

We can write this as:

E = (1/(4πε0)) * [(Q/x^2) - (Q/(d-x)^2)] * ê

where ê is a unit vector in the direction of the electric field. Note that the electric field is a vector quantity, so we need to specify both its magnitude and direction.

The direction of the electric field is from charge A to charge B, since this is the direction in which a positive test charge would be pushed. The magnitude of the electric field depends on the distances x and (d-x), as well as the charges Q and -Q.

We can simplify the expression for E by using the fact that x + (d-x) = d:

E = (1/(4πε0)) * [(Q/d^2) * (2x - d)] * ê

This shows that the electric field at point P is proportional to the difference between the distances of the point from the two charges. If the point P is equidistant from the two charges (i.e. x = d/2), then the electric field is zero. This is because the electric fields due to the two charges cancel out at this point.

If we move the point P towards one of the charges, the electric field due to that charge becomes stronger, while the field due to the other charge becomes weaker. The electric field at point P is strongest when it is very close to one of the charges, and weakest when it is in the middle of the two charges.

Overall, the electric field due to two point charges is a simple example of how electric fields can be added together to find the total field at a point. This principle can be extended to more complex systems of charges, by adding up the electric fields due to each individual charge.
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Two point charges A and B, having charges +Q and –Q respectively...
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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)Fb)9F/16c)16F/9d)4F/3Correct answer is option 'B'. Can you explain this answer?
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