A positively charged sphere with a charge of 8Q is separated from a negatively charged sphere -2Q by a distance r. There is an attractive force F exerted on each sphere. The spheres briefly touch each other and move to the original distance r. What is the new force on each sphere in terms of F

Respuesta :

Answer:

Explanation:

For the first case , the expression for electrostatic force can be given by the following .

F = K x 8Q x 2Q / r² where k is a constant .

F = K 16 Q² / r²

When they touch , some charge is neutralized . Net charge remaining

= 8Q - 2 Q = 6 Q

Charge on each sphere = 6Q/2 = 3 Q .

Force between them

F₁ = k 3Q x 3 Q / r² = k 9 Q² / r²

F₁ / F = 9 / 16

F₁ = 9 F / 16 .

The new force on each sphere will be "[tex]\frac{9}{16} F[/tex]".

We have,

  • [tex]Q_1 = +8Q[/tex]
  • [tex]Q_2 = -2 Q[/tex]

Now,

The initial force between them will be:

→ [tex]|\vec{F}| = \frac{1}{4 \pi \epsilon_0 } \frac{Q_1 Q_2}{r^2}[/tex]

        [tex]= \frac{1}{4 \pi \epsilon_0 } \frac{8Q\times (2Q)}{r^2}[/tex]

        [tex]= \frac{16 Q^2}{4 \pi \epsilon r^2}[/tex]

When the separated, the charge on each become

→ [tex]Q'= \frac{Q_1+Q_2}{2}[/tex]

       [tex]= \frac{+8Q-2Q}{2}[/tex]

       [tex]= \frac{6Q}{2}[/tex]

       [tex]= 3Q[/tex]

For "Q" between them,

→ [tex]F' = \frac{1}{4 \pi \epsilon_0 } \frac{Q'\times Q'}{r^2}[/tex]

       [tex]= \frac{1}{4 \pi \epsilon_0 }\frac{3Q\times 3Q}{r^2}[/tex]

       [tex]= \frac{1}{4 \pi \epsilon_0} \frac{9Q^2}{r^2}[/tex]

hence,

The new force will be:

→ [tex]\frac{F'}{F} = \frac{1}{4 \pi \epsilon_0} \frac{9 Q^2}{r^2}\times \frac{4 \pi \epsilon r^2}{16 Q^2}[/tex]

        [tex]= \frac{9}{16}[/tex]

    [tex]F' = \frac{9}{16} F[/tex]

Thus the above answer is correct.  

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