Question

If electrical force between two charges is $ \text{2}00\text{ N} $ and we increase $ \text{1}0% $ charge on one of the charges and decrease $ \text{1}0% $ charge on the other, then electrical force between them for the same distance becomes

• A. $ \text{2}00\text{ N} $
• B. 202 N
• C. 198 N
• D. 199 N

Answer 1

Answer: (C)

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 $