A) \[\text{2}00\text{ N}\]
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 \[F=\frac{1}{4\pi {{\varepsilon }_{0}}}.\frac{{{q}_{1}}{{q}_{2}}}{{{r}^{2}}}=200N\] ? (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 \[F'=200\times \frac{99}{100}\Rightarrow F'=198N\]
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