A) \[9\times {{10}^{-6}}C,\,-3\times {{10}^{-6}}C\]
B) \[1\times {{10}^{-6}}C,\,-3\times {{10}^{-6}}C\]
C) \[-3\times {{10}^{-6}}C,\,\,\,2\times {{10}^{-6}}C\]
D) \[1\times {{10}^{-6}}C,-\,2\times {{10}^{-6}}C\]
Correct Answer: B
Solution :
Given that \[{{F}_{1}}=0.108\,N\] \[r=0.5\,m\] \[{{F}_{1}}=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{{{q}_{1}}{{q}_{2}}}{{{r}^{2}}}\] \[0.108=9\times {{10}^{9}}\,\,\frac{{{q}_{1}}{{q}_{2}}}{{{(0.5)}^{2}}}\] \[{{q}_{1}}{{q}_{2}}=\frac{0.25\times 0.108}{9\times {{10}^{9}}}=\frac{108\times 25\times {{10}^{-5}}}{9\times {{10}^{9}}}\] \[{{q}_{1}}{{q}_{2}}=3\times {{10}^{-12}}{{C}^{2}}\] ... (i) and \[0.036=9\times {{10}^{9}}\frac{{{\left( \frac{{{q}_{1}}-{{q}_{2}}}{2} \right)}^{2}}}{{{(0.5)}^{2}}}\] \[=\frac{0.036\times 4\times 0.25}{9\times {{10}^{9}}}={{({{q}_{1}}-{{q}_{2}})}^{2}}\] \[\frac{4\times 4\times 25\times {{10}^{5}}}{{{10}^{9}}}{{({{q}_{1}}-{{q}_{2}})}^{2}}\] \[4\times {{10}^{-12}}={{({{q}_{1}}-{{q}_{2}})}^{2}}\] \[2\times {{10}^{-6}}={{q}_{1}}-{{q}_{2}}\] ... (iii) Now \[{{({{q}_{1}}+{{q}_{2}})}^{2}}={{({{q}_{1}}-{{q}_{2}})}^{2}}+4\,{{q}_{1}}{{q}_{2}}\] \[=4\times {{10}^{-12}}+4\times 3\times {{10}^{-12}}\] \[=16\times {{10}^{-12}}\] \[{{q}_{1}}+{{q}_{2}}=4\times {{10}^{-6}}\] ... (iv) From Eqs. (iii) and (iv), we get \[{{q}_{1}}=-3\times {{10}^{-6}}\] and \[{{q}_{2}}=1\times {{10}^{6}}\]
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