A) 200 N
B) 202 N
C) 198 N
D) 199 N
Correct Answer: C
Solution :
Let two charges are \[{{q}_{1}}\] and \[{{q}_{2}}\] and r is the distance between them, then electrical force \[F=\frac{1}{4\pi {{\varepsilon }_{0}}}.\frac{{{q}_{1}}{{q}_{2}}}{{{r}^{2}}}=200\,N\] ? (i) If \[{{q}_{1}}\] is increased by 10%, then \[q_{1}^{}=\frac{110}{100}{{q}_{1}}\] and \[{{q}_{2}}\] is decreased by 10%, then \[q_{2}^{}=\frac{90}{100}\,{{q}_{2}}\] Then, electrical force between them, \[F=\frac{1}{4\pi {{\varepsilon }_{0}}}\times \frac{q_{1}^{}q_{2}^{}}{{{r}^{2}}}\] \[F=\frac{\frac{1}{4\pi {{\varepsilon }_{0}}}\times \frac{110}{100}{{q}_{1}}\times \frac{90}{100}{{q}_{2}}}{{{r}^{2}}}\] ?. (ii) From Eqs. (i) and (ii), we get \[F=200\times \frac{99}{100}\] \[\Rightarrow \] \[F=198\,N\]
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