A) F/2
B) F/8
C) F
D) Zero
Correct Answer: C
Solution :
From Coulombs law, the force of attraction between two charged particles\[(q)\], kept at distance r apart is when two identical spheres are brought in contact, charge on them is equalized, hence total charge on\[C\]is equally shared when brought in contact with sphere\[B\]having a charge\[q\]. Therefore, charge on\[B\]and\[C\]is\[\frac{q}{2}\] . From Coulombs law, the force on\[C\]is \[{{F}_{C}}=\frac{q\times q/2}{4\pi {{\varepsilon }_{0}}{{(r/2)}^{2}}}-\frac{(q/2)(q/2)}{4\pi {{\varepsilon }_{0}}{{(r/2)}^{2}}}\] \[=\frac{qq}{4\pi {{\varepsilon }_{0}}{{r}^{2}}}(2-1)=F\] Note: The force will be opposite because\[A\]and\[B\]spheres will repel the third sphere.You need to login to perform this action.
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