Solve the following :

Question:

Two charged particles placed at a separation of $20 \mathrm{~cm}$ exert $20 \mathrm{~N}$ of Coulomb force on each other. What will be the force if the separation is increased to $25 \mathrm{~cm}$ ?

Solution:

$F=\frac{k q_{1} q_{2}}{r^{2}}$

$F \propto \frac{1}{r^{2}}$

$\frac{F}{F^{\prime}}=\frac{r^{\prime 2}}{r^{2}}$

$F^{\prime}=\frac{20 \mathrm{x}(0.20)^{2}}{(0.25)^{2}}$

$F^{\prime}=12.8 \mathrm{~N}$

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