NEET-XII-Physics
04: The Forces
- #8Two charged particles placed at a separation of 20 cm exert 20 N of Coulomb force on each other. What will be the force of the separation is increased to 25 cm?Ans : Two charged particles placed at a separation of 20 cm exert 20 N of Coulomb force on each other.
So, `` {F}_{1}=\frac{1}{4\,\mathrm{\,\pi \,}{\in }_{0}}·\frac{{q}^{2}}{{r}_{1}^{2}}``
Also, `` {F}_{2}=\frac{1}{4\,\mathrm{\,\pi \,}{\in }_{0}}·\frac{{q}^{2}}{{r}_{2}^{2}}``
According to the question, we have:
`` \frac{{F}_{2}}{{F}_{1}}=\frac{{r}_{1}^{2}}{{r}_{2}^{2}}``
`` =\frac{20\times 20}{25\times 25}=\frac{16}{25}``
`` \therefore {F}_{2}=\frac{16}{25}\times {F}_{1}``