Question 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 w

When sphere A is touched with an uncharged sphere

C, \frac{q}{2}

amount of charge from A will transfer to sphere C. Hence, charge on each of the spheres, A and C, is

\frac{q}{2}

When sphere C with charge

\frac{q}{2}

. is brought in contact with sphere B with charge q, total charges on the system will divide into two equal halves given as,

Each sphere will share each half. Hence, charge on each of the spheres, C and B, is

\frac{3 q}{4}

Force of repulsion between sphere A having charge

\frac{q}{2}

and sphere B having charge =

\frac{3 q}{4}

= 9 \times 10^{9} \times \frac{3 \times {(6.5 \times 10^{- 7})}^{2}}{8 \times {(0.5)}^{2}}

Question 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 w- Hosting Precise

Posted on 2nd Oct 2021 06:31:23 in

Ans:-

Distance between the spheres, A and B

r = 0.5 m

Initially, the charge on each sphere

$q = 6.5 \times 10^{- 7} C$

When sphere A is touched with an uncharged sphere $C, \frac{q}{2}$ amount of charge from A will transfer to sphere C. Hence, charge on each of the spheres, A and C, is $\frac{q}{2}$ .

When sphere C with charge $\frac{q}{2}$ . is brought in contact with sphere B with charge q, total charges on the system will divide into two equal halves given as,

$\frac{\frac{q + q}{2}}{2} = \frac{3 q}{4}$

Each sphere will share each half. Hence, charge on each of the spheres, C and B, is $\frac{3 q}{4}$ .

Force of repulsion between sphere A having charge $\frac{q}{2}$ and sphere B having charge = $\frac{3 q}{4}$

$= \frac{\frac{q}{2} \times \frac{3 q}{4}}{4 π \in o r^{2}}$

$= \frac{3 q^{2}}{8 \times 4 π \in o r^{2}}$

$= 9 \times 10^{9} \times \frac{3 \times {(6.5 \times 10^{- 7})}^{2}}{8 \times {(0.5)}^{2}}$

$= 5.703 \times 10^{- 3} N$

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