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Electrostatics — Physics STD 11 Science — Question

Maharashtra BoardEnglish MediumSTD 11 SciencePhysicsElectrostatics4 Marks
Question
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 (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 (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 (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 comparison to the separation between their centres.
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Answer

 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

$F =\frac{1}{4 \pi \varepsilon_0} \frac{ qq ^{\prime}}{ r ^2}$
Neglecting the sizes of spheres, A and B in comparison to r. 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

$F ^{\prime}=\frac{1}{4 \pi \varepsilon_0} \frac{( q / 2)\left( q ^{\prime} / 2\right)}{( r / 2)^2}=\frac{1}{4 \pi \varepsilon_0} \frac{\left( qq ^{\prime}\right)}{ r ^2}= F$
Thus, the electrostatic force on A, due to B, remains unaltered.

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