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Standard XII

Physics

Potential Energy - External Viewpoint

The force bet...

Question

The force between two charges 0.06m apart is 5N. If each charge is moved towards the other by 0.01m, then the force between them will become

170 N

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11.25N

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45N

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22.50 N

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Solution

The correct option is B 11.25N
$\frac{F_{2}}{F_{1}} = {(\frac{d_{1}}{d_{2}})}^{2} = {(\frac{0.06}{0.04})}^{2}$
$F_{2} = \frac{9}{4} F_{1} = \frac{9}{4} \times 5 = \frac{45}{4} = 11.25 N$

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The force between two charges 0.06m apart is 5N. If each charge is moved towards the other by 0.01m, then the force between them will become

The correct option is B 11.25NF2F1=(d1d2)2=(0.060.04)2 F2=94F1=94×5=454=11.25N

The correct option is B 11.25N
$\frac{F_{2}}{F_{1}} = {(\frac{d_{1}}{d_{2}})}^{2} = {(\frac{0.06}{0.04})}^{2}$
$F_{2} = \frac{9}{4} F_{1} = \frac{9}{4} \times 5 = \frac{45}{4} = 11.25 N$