Two similarly charged spheres A and B separated by a distance 'r' with force of 2×10^-5N. A third identical uncharged sphere C is touched to A and then placed between a mid point of A and B. Calculate net electrostatic force on C?

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Answer 1 · 2018-05-04T07:03:29

$Please have a look at the fallowing explanation.$ $"Please have a look at the fallowing explanation."$

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Coulomb's law can help you calculate the force that two electric charges apply to each other.
$F = k \cdot \frac{q_{1} \cdot q_{2}}{r^{2}}$ $F=k*(q_1*q_2)/r^2$
$2 \cdot 10^{- 5} = k \cdot \frac{q \cdot q}{r^{2}} = k \cdot \frac{q^{2}}{r^{2}}$ $color(red)(2*10^-5)=k*(q*q)/r^2=color(red)(k*q^2/r^2)$
The force is directly proportional to the product of the electric charges.
The force is inversely proportional to the square of the distance between the electric charges.

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The new electric charge distribution of the A and C spheres is shown in Figure 3.
Figure 4 shows the new load sequence.

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C is pushed by both A and B.
$F_{1} = k \cdot \frac{\frac{q}{2} \cdot \frac{q}{2}}{{(\frac{r}{2})}^{2}} = k \cdot \frac{\frac{q^{2}}{4}}{\frac{r^{2}}{4}} = k \cdot \frac{q^{2}}{r^{2}} = {2.10}^{- 5}$ $F_1=k*(q/2*q/2)/(r/2)^2=k*(q^2/4)/(r^2/4)=color(red)(k*q^2/r^2=2.10^-5)$
$F_{2} = k \cdot \frac{q \cdot \frac{q}{2}}{{(\frac{r}{2})}^{2}} = k \cdot \frac{\frac{q^{2}}{2}}{\frac{r^{2}}{4}} = 2 k \cdot \frac{q^{2}}{4} = 2 \cdot 2 \cdot 10^{- 5} = 4 \cdot 10^{- 5}$ $F_2=k*(q*q/2)/(r/2)^2=k*(q^2/2)/(r^2/4)=2color(red)(k*q^2/4)=color(red)(2*2*10^-5=4*10^-5)$
Now we must find the vector sum of $F_{1}$ $F_1""$ and $F_{2}$ $" " F_2$ .
$\to R = {\to F}_{1} + {\to F}_{2}$ $vec R=vec F_1 +vec F_2$
magnitude of net electrostatic force over C is:
$- 4 \cdot 10^{- 5} + 2 \cdot 10^{- 5} = - 2 \cdot 10^{- 5} Newtons$ $-4*10^-5+2*10^-5=-2*10^-5" Newtons"$
The direction to B is considered positive.
-The direction of net electrostatic force is toward A.