
Two spherical conductors B and C 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 having the same radius as that of B put uncharged, is brought in contact with B, then brought in contact with C and finally removed away from both. The new force of repulsion between B and C is
A. $\dfrac{F}{4}$
B. $\dfrac{{3F}}{4}$
C. $\dfrac{F}{8}$
D. $\dfrac{{3F}}{8}$
Answer
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Hint:Conductors are composed of the mobile charge carriers. In case of metallic conductors, these charge carriers are known as electrons. In metals, the valence electrons are parted away from their atoms and thus, are free to move. When two identical metallic surfaces brought in contact then charge on them are equalised due to the flow of free electrons.
Complete step by step answer:
Consider two sphere B and C both having the same radius (R). Put both the sphere at x distance or we can say that the distance between the two spheres is x. as shown below:
Both having the same charge say ‘q’. Then force form sphere B to C can be expressed with the following formula:
$F = k\dfrac{{{q^2}}}{{{x^2}}}$ (i)
Now, third sphere A has been introduced. When the third sphere comes in contact with B, then new charge B is$\dfrac{q}{2}$.
And charge on the third conductor is$\dfrac{q}{2}$. When third conductor come in contact with C sphere then total charge$b{f_{BC}}$ becomes$q + \dfrac{q}{2} = \dfrac{3}{2}q$. Charge will equally distributed, so charge on $C = \dfrac{3}{4}q$
Now, new force
$F_{BC}^1 = k\dfrac{{\dfrac{q}{2}\times \dfrac{3}{4}q}}{{{x^2}}}$
Put value of${F_{BC}}$ in above we get,
$\therefore{F_{BC}^1} = \dfrac{3}{8}{F_{BC}}$
Hence, the correct answer is Option D.
Note: It should be noted that the charge is the fundamental property of the form of matter that exhibits the electrostatic attraction or repulsion in the presence of other matter. Charges may be positive or negative. Charge will be distributed equally that is why in the solution, whenever a third conductor is added then charge distributed equally.
Complete step by step answer:
Consider two sphere B and C both having the same radius (R). Put both the sphere at x distance or we can say that the distance between the two spheres is x. as shown below:
Both having the same charge say ‘q’. Then force form sphere B to C can be expressed with the following formula:
$F = k\dfrac{{{q^2}}}{{{x^2}}}$ (i)
Now, third sphere A has been introduced. When the third sphere comes in contact with B, then new charge B is$\dfrac{q}{2}$.
And charge on the third conductor is$\dfrac{q}{2}$. When third conductor come in contact with C sphere then total charge$b{f_{BC}}$ becomes$q + \dfrac{q}{2} = \dfrac{3}{2}q$. Charge will equally distributed, so charge on $C = \dfrac{3}{4}q$
Now, new force
$F_{BC}^1 = k\dfrac{{\dfrac{q}{2}\times \dfrac{3}{4}q}}{{{x^2}}}$
Put value of${F_{BC}}$ in above we get,
$\therefore{F_{BC}^1} = \dfrac{3}{8}{F_{BC}}$
Hence, the correct answer is Option D.
Note: It should be noted that the charge is the fundamental property of the form of matter that exhibits the electrostatic attraction or repulsion in the presence of other matter. Charges may be positive or negative. Charge will be distributed equally that is why in the solution, whenever a third conductor is added then charge distributed equally.
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