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Coulomb's Law - Mathematical Form

A positive ch...

Question

A positive charged particle is kept on the plane surface. There are two points X and Y lies on the plane surface as shown above. If the point Y is three times as far away from Q as point X. What is the ratio of the electric force that would act on a small charge placed at point Y compared to the charge placed at point X?

A

1 : 9

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B

1 : 3

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C

1 : 1

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D

3 : 1

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E

9 : 1

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Solution

The correct option is B $1 : 9$
Given : $Q X = r$ $Q Y = 3 r$
Let the positive charge at point Q be $q$ .
Electric field due to $q$ at point X $E_{x} = \frac{k q}{r^{2}}$

Electric field due to $q$ at point Y $E_{y} = \frac{k q}{(3 r)^{2}} = \frac{k q}{9 r^{2}}$
$⟹$ $\frac{E_{y}}{E_{x}} = \frac{1}{9}$

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