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Electric Charges and Fields — Physics STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 SciencePhysicsElectric Charges and Fields3 Marks
Question
A charged metallic sphere A is suspended by a nylon thread. Another charged metallic sphere B held by an insulating
Image
handle is brought close to A such that the distance between their centres is 10 cm, as shown in Fig. 1.4(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 Fig. 1.4(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 Fig. 1.4(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.
✓

Answer

Let the original charge on sphere A be $q$ and that on B be $q^{\prime}$. 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{q q^{\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 $q^{\prime} / 2$. 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(q q^{\prime}\right)}{r^2}=F
$
Thus the electrostatic force on A, due to B, remains unaltered.

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