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  4. Each of the two point charges are doubled and their distance is halved. Force of interaction becomes n times, where n is

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Q. Each of the two point charges are doubled and their distance is halved. Force of interaction becomes n times, where n is

BVP MedicalBVP Medical 2008

Solution:

From the formula, $ F=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{{{q}_{1}}{{q}_{2}}}{{{r}^{2}}} $ $ F=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{2{{q}_{1}}\times 2{{q}_{2}}}{{{(r/2)}^{2}}} $ $ =\frac{4}{1/4}F $ $ \Rightarrow $ $ F=16F $ $ \Rightarrow $ $ nF=16F $ $ \therefore $ $ n=16 $