Ncert Solutions Physics Class 12th Physics Ncert Solutions Class 12th Physics Class 12th Electric Charges and Fields

1.13 Suppose the spheres A and B in Exercise 1.12 have identical sizes. A third sphere of the same size but uncharged is brought in contact with the first, then brought in contact with the second, and finally removed from both. What is the new force of repulsion between A and B?

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1 Answer

Answered by

alok kumar singh | Contributor-Level 10

3 months ago

1.13 When the third uncharged sphere C is brought in contact with the sphere A, then the charge is shared and becomes half. Then

$q_{A}$ = $\frac{q}{2}$ and = $q_{C}$ = $\frac{q}{2}$

When the charged sphere C is brought in contact with charged sphere B, the charge between both the sphere is shared and becomes half

$q_{B}, q_{C} = \frac{1}{2}$ (q + = $\frac{q}{2})$ $\frac{3 q}{4}$

Hence the force of repulsion between sphere A and B can be given as

F =
$\frac{1}{4 π ?_{0}}$ $\times \frac{q_{1} q_{2}}{r^{2}}$ , where $?_{0}$ = Permittivity of free space = 8.854 $\times 10^{- 12}$ $C^{2} N^{- 1}$ $m^{- 2}$ = = $\frac{1}{4 \times π \times 8.854 \times 10^{- 12}}$ = $\times \frac{(\frac{q}{2} \times \frac{3 q}{4})}{{0.5}^{2}}$ = $\frac{1}{4 \times π \times 8.854 \times 10^{- 12}}$ $\times \frac{3 \times q^{2}}{{8 \times 0.5}^{2}}$
= $\frac{1}{4 \times π \times 8.854 \times 10^{- 12}}$ $\times \frac{3 \times {(6.5 \times 10^{- 7})}^{2}}{{8 \times 0.5}^{2}}$ = 5.695 $\times 10^{- 3}$ N

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