NEET-XII-Physics
29: Electric Field and Potential
- #20Ten positively-charged particles are kept fixed on the x-axis at points x = 10 cm, 20 cm, 30 cm, ...., 100 cm. the first particle has a charge 1.0 × 10-8 C, the second 8 × 10-8 C, the third 27 × 10-8 C and so on. The tenth particle has a charge 1000 × 10-8 C. Find the magnitude of the electric force acting on a 1 C charge placed at the origin.Ans : By Coulomb's Law, force (F) on charge q due to one charge,
`` F=\frac{1}{4\,\mathrm{\,\pi \,}{\epsilon }_{0}}\frac{{q}_{1}q}{{r}^{2}}``
So, net force due to ten charges,
`` F={F}_{1}+{F}_{2}+{F}_{3}+.....+{F}_{10}``
`` =9\times {10}^{9}\times \left[\frac{1.0}{{\left(0.10\right)}^{2}}+\frac{8}{{\left(.20\right)}^{2}}+\frac{27}{{\left(.30\right)}^{2}}+......\frac{1000}{{\left(1.0\right)}^{2}}\right]{10}^{-8}``
`` =\frac{9\times {10}^{9}\times {10}^{-8}}{{10}^{-2}}\left[1+2+3+......10\right]``
`` =9\times {10}^{3}\times 55``
`` =4.95\times {10}^{5}\,\mathrm{\,N\,}``
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