The force of repulsion between 2 Point charges placed 16cm apart in vaccum is 7.5×10^-10 B. What will be force between them if they are placed in a medium of dielectric constant k=2.5?

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Arun · Answer

Dear Arushi r = 1.6 cm F(Vac) = 7.5 * 10 &ndash;10 N K = 2.5 F(vac)/ F(medium) = &epsilon;/&epsilon; 0 = K Now F(medium) = F(vac)/ K = 3 * 10 &ndash;10 N Regards Arun (askIITians forum expert)

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The force of repulsion between 2 Point charges placed 16cm apart in vaccum is 7.5×10^-10 B. What will be force between them if they are placed in a medium of dielectric constant k=2.5?

The force of repulsion between 2 Point charges placed 16cm apart in vaccum is 7.5×10^-10 B. What will be force between them if they are placed in a medium of dielectric constant k=2.5?

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The force of repulsion between 2 Point charges placed 16cm apart in vaccum is 7.5×10^-10 B. What will be force between them if they are placed in a medium of dielectric constant k=2.5?

The force of repulsion between 2 Point charges placed 16cm apart in vaccum is 7.5×10^-10 B. What will be force between them if they are placed in a medium of dielectric constant k=2.5?

1 Answers

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