Physics, asked by BrainlyQueenPihu, 11 months ago

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?

Answers

Answered by itzdevilqeen
19

Answer:

The force of attraction between the two sphere is \large \:5.703 \:× \:10^{-3}N

Explanation:

Distance between the spheres,

\large \:A \:and \:B, \:r \:= \:0.5m

Initially, the charge on each sphere,

\large \:q \:= \:6.5 \:× \:10^{-7} \:C

When sphere A is touched with an uncharged sphere C, amount of charge from A will transfer to sphere C. Hence, charge on each of the sphere, A and C,

\large \frac{q}{2}

When sphere C with charge

\large \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,

\large \frac{\frac{q}{2} + a}{2} \:= \frac{3q}{4}

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

\large \frac{3q}{4}

Force of repulsion between sphere A having charge

\large \frac{q}{2}

and sphere B having charge

\large \frac{3q}{4}

\large \:= \frac{\frac{q}{2} × \frac{3q}{4}}{4πEr^{2}}

\large \:= \frac{3q^{2}}{8 × 4πEr^{2}}

\large \:= \:9 \:× \:10^{9} \:× \frac{3 × ( 6.5 × 10^{-1} )^{2}}{8 × (0.5)^{2}}

\large \:= \:5.703. \:× \:10^{-3}N

Therefore, the force of attraction between the two sphere is \large \:5.703 \:× \:10^{-3}N

Answered by Anonymous
65

Explanation:

\Large{\red{\underline{\underline{\sf{\blue{Solution:}}}}}}

 ‎

 ‎ ‎ ‎ ‎ ‎ ‎

\hookrightarrow Distance between the spheres, A and B, r = 0.5m

\hookrightarrow Initially, the charge on each sphere, \sf q\:=\:6.5\times 10^{-7}C

 ‎ ‎ ‎ ‎ ‎ ‎ ‎

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

\sf \dfrac{q}{2}

 ‎ ‎ ‎ ‎ ‎ ‎ ‎

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

 ‎ ‎ ‎ ‎ ‎ ‎ ‎

\sf \implies\: \dfrac{\dfrac{q}{2}+2}{2}\:=\: \dfrac{3q}{4}

 ‎ ‎ ‎ ‎ ‎ ‎ ‎

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

 ‎ ‎ ‎ ‎ ‎ ‎ ‎

= \sf \dfrac{\dfrac{q}{2}\times \dfrac{3q}{4}}{4\pi\,\epsilon_{\circ}r^2}\:=\: \dfrac{3q^2}{8\times 4\pi\,\epsilon_{\circ}r^2}

 ‎ ‎ ‎ ‎ ‎ ‎ ‎

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

 ‎ ‎ ‎ ‎ ‎ ‎ ‎

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

 ‎ ‎ ‎ ‎ ‎ ‎ ‎

\hookrightarrow Hence, the force of attraction between the two spheres is \sf{\purple{5.703\times 10^{-3}N}}

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