NEET-XII-Physics
29: Electric Field and Potential
- #16Three equal charges, 2.0 × 10-6 C each, are held at the three corners of an equilateral triangle of side 5 cm. Find the Coulomb force experienced by one of the charges due to the other two.Ans : Since all the charges are of equal magnitude, the force on the charge at A due to the charges at B and C will be of equal magnitude. (As shown in the figure)
That is, `` {F}_{\,\mathrm{\,BA\,}}={F}_{\,\mathrm{\,CA\,}}=F\left(\,\mathrm{\,say\,}\right)``
The horizontal components of force cancel each other and the net force on the charge at A,
`` F\text{'}={F}_{\,\mathrm{\,BA\,}}\,\mathrm{\,sin\,}\theta +{F}_{\,\mathrm{\,CA\,}}\,\mathrm{\,sin\,}\theta ``
`` \,\mathrm{\,F\,}\text{'}=2F\,\mathrm{\,sin\,}\theta ``
`` ``
Given: r = 5 cm =0.05 m
By Coulomb's Law, force,
`` F=\frac{1}{4\,\mathrm{\,\pi \,}{\epsilon }_{0}}\frac{{q}_{1}{q}_{2}}{{r}^{2}}``
`` F\text{'}=\frac{2\times 9\times {10}^{9}\times {\left(2\times {10}^{-6}\right)}^{2}\times {\displaystyle \,\mathrm{\,sin\,}60°}}{{\left(0.05\right)}^{2}}``
`` F\text{'}=\frac{2\times 9\times {10}^{9}\times {\left(2\times {10}^{-6}\right)}^{2}\times {\displaystyle \frac{\sqrt{3}}{2}}}{{\left(0.05\right)}^{2}}``
F' = 24.9 N
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