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Standard XII

Physics

Charging by Induction

Two particles...

Question

Two particles having charges $q_{1}$ and $q_{2}$ when kept at a certain distance, exert a force F on each other. If the distance between the two charges is reduced to half then the force exerted on each other would be:

2 F

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4 F

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8 F

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16 F

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Solution

The correct option is B $4 F$
Let the force be $F = \frac{k q_{1} q_{2}}{r^{2}}$ ....(1)
so distance is reduced to $r / 2$
so $F_{1} = \frac{k q_{1} q_{2}}{(\frac{r}{2})^{2}} = \frac{4 k q_{1} q_{2}}{r^{2}} = 4 F$ by using equation 1.

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Two particles having charges q1 and q2 when kept at a certain distance, exert a force F on each other. If the distance between the two charges is reduced to half then the force exerted on each other would be:

The correct option is B 4 FLet the force be F=kq1q2r2 ....(1)so distance is reduced to r/2so F1=kq1q2(r2)2=4kq1q2r2=4F by using equation 1.

Two particles having charges $q_{1}$ and $q_{2}$ when kept at a certain distance, exert a force F on each other. If the distance between the two charges is reduced to half then the force exerted on each other would be:

The correct option is B $4 F$
Let the force be $F = \frac{k q_{1} q_{2}}{r^{2}}$ ....(1)
so distance is reduced to $r / 2$
so $F_{1} = \frac{k q_{1} q_{2}}{(\frac{r}{2})^{2}} = \frac{4 k q_{1} q_{2}}{r^{2}} = 4 F$ by using equation 1.