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Home  >>  AIMS  >>  Class12  >>  Physics  >>  Electric Charges and Fields
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A charged metallic sphere A is suspended by a nylon thread. Another charged metallic sphere B held by an insulating handle is brought close to A such that the distance between their centres is 10 cm, as shown in Figure 1.20. (a). The resulting repulsion of A is noted (for example, by shining a beam of light and measuring the deflection of its shadow on a screen). Spheres A and B are touched by uncharged spheres C and D respectively, as shown in Figure 1.20. (b). C and D are then removed and B is brought closer to A to a distance of 5.0 cm between their centres, as shown in Figure 1.20 (c). What is the expected repulsion of A on the basis of Coulomb’s law? Spheres A and C and spheres B and D have identical sizes. Ignore the sizes of A and B in comparison to the separation between their centres

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Let the original charge on sphere A be q and that on B be q.
At a distance r between their centres, the magnitude of the electrostatic force on each is given by
$F=\large\frac{1}{4 \pi \in_0} \frac{qq'}{r^2}$
The sizes of spheres A and B in comparison to r are neglected. When an identical but uncharged sphere C touches A, the charges redistribute on A and C and, by symmetry, each sphere carries a charge q/2.
Similarly, after D touches B, the redistributed charge on each is q/2. Now, if the separation between A and B is halved, the magnitude of the electrostatic force on each is
$F'= \large\frac{1}{4 \pi \in_0} \frac{(q/2)(q'/2)}{(r/2)^2}$
$\quad= \large\frac{1}{4 \pi \in_0 } \frac{(qq')}{r^2}$$=F$
Thus the electrostatic force on A, due to B, remains unaltered
Hence B is the correct answer.
answered May 29, 2014 by meena.p
 

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