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NEET-XII-Physics
29: Electric Field and Potential
- #17Four equal charges of 2.0 × 10-6 C each are fixed at the four corners of a square of side 5 cm. Find the Coulomb’s force experienced by one of the charges due to the other three.Ans : Given,
Magnitude of the charges, `` q=2\times {10}^{-6}\,\mathrm{\,C\,}``
Side of the square, `` a=5\,\mathrm{\,cm\,}=0.05\,\mathrm{\,m\,}``
By Coulomb's Law, force,
`` F=\frac{1}{4\,\mathrm{\,\pi \,}{\epsilon }_{0}}\frac{{q}_{1}{q}_{2}}{{r}^{2}}``
So, force on the charge at A due to the charge at B,
`` {\stackrel{\to }{F}}_{\,\mathrm{\,BA\,}}=\frac{9\times {10}^{9}\times {\left(2\times {10}^{-6}\right)}^{2}}{{\left(0.05\right)}^{2}}``
`` =\frac{9\times {10}^{9}\times 4\times {10}^{-12}}{25\times {10}^{-4}}``
`` =14.4\,\mathrm{\,N\,}``
`` ``
Force on the charge at A due to the charge at C,
`` {\stackrel{\to }{F}}_{\,\mathrm{\,CA\,}}=\frac{9\times {10}^{9}\times {\left(2\times {10}^{-6}\right)}^{2}}{{\left(\sqrt{2}\times 0.05\right)}^{2}}``
`` =\frac{9\times {10}^{9}\times 4\times {10}^{-12}}{25\times 2\times {10}^{-4}}``
`` =7.2\,\mathrm{\,N\,}``
Force on the charge at A due to the charge at D,
`` {\stackrel{\to }{F}}_{\,\mathrm{\,DA\,}}={\stackrel{\to }{F}}_{\,\mathrm{\,BA\,}}``
The resultant force at A, F'= `` {\stackrel{\to }{F}}_{\,\mathrm{\,BA\,}}+{\stackrel{\mathit{\to }}{\mathit{F}}}_{\,\mathrm{\,CA\,}}+{\stackrel{\mathit{\to }}{\mathit{F}}}_{\,\mathrm{\,DA\,}}``
The resultant force of `` {\stackrel{\to }{F}}_{\,\mathrm{\,DA\,}}\,\mathrm{\,and\,}{\stackrel{\to }{F}}_{\,\mathrm{\,BA\,}}`` will be `` \sqrt{2}{F}_{\,\mathrm{\,BA\,}}`` in the direction of `` {\stackrel{\mathit{\to }}{F}}_{\,\mathrm{\,CA\,}}``. Hence, the resultant force,
`` F\text{'}=14.4\sqrt{2}+7.2``
`` =27.56\,\mathrm{\,N\,}``
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