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Two point charges $\;q_{1}$ and $\;q_{2}$ are placed at a certain distance d from each other . The interacting force between them is F . Now both charges kept in oil of relative permittivity 5 . What is the new value of force between them

$(a)\;\large\frac{F}{2}\qquad(b)\;\large\frac{F}{4}\qquad(c)\;\large\frac{F}{5}\qquad(d)\;\large\frac{F}{25}$

1 Answer

Answer : (c) $\;\large\frac{F}{5}$
Explanation :
$F=\large\frac{1}{4\pi \in_{0}}\;\large\frac{q_{1} q_{2}}{d^2}$
$F^{|}=\large\frac{1}{4\pi \in}\;\large\frac{q_{1} q_{2}}{d^2}$
$F^{|}=\large\frac{1}{5}\;\large\frac{1}{4\pi \in_{0}}\;\large\frac{q_{1} q_{2}}{d^2}$
$=\large\frac{F}{5}\;.$
answered Feb 6, 2014 by yamini.v
 

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