Practicing Success
Practicing Success
CUET Preparation Today
Read the passage and answer the question. Two spherical conductors A and B having equal radii and carrying equal charges in them repel each other with a force F, when kept apart at some distance. A third spherical conductor P having the same radius as that of A (or B) but uncharged, is brought in contact with A and then removed away from both. |
If the third spherical conductor P is then brought in contact with B and moved away, what is the new force of repulsion between A and B ? |
F/2 F/4 3F/8 3F/16 |
3F/8 |
Let A and B have charge Q each initially and are seperated by r distance.}
$\text{From Coulomb's law, force between them is }F = \frac{kQ^2}{r^2}$
$\text{When an identical uncharged spherical conductor P is brought in contact with charged sphere A.}$
$\text{By symmetry, the total charge will be shared equally among them, as both have equal radii.}$
$\text{Charge on P and A are } = \frac{Q}{2}$
$\text{Now, when the conductor P is brought in contact with B. }$
$\text{They will also share equal charge among themselves, as both have equal radii.}$
$\text{Charge on P and B }= \frac{Q+\frac{Q}{2}}{2}= \frac{3Q}{4}$
$\text{New Force is }F'= \frac{kQ/2 \times 3Q/4}{r^2} = \frac{3F}{8}$
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