Question

Electrical force between two charged spheres is $200\text{N}$.If we increase $10\%$ charge on one of the charged sphere and decrease $20\%$ charge in the other,then the electrical force between them for the same distance becomes

• A. $156\text{N}$
• B. $167\text{N}$
• C. $176\text{N}$
• D. $216\text{N}$

Answer 1

The correct option is C $176\text{N}$

From coulumb's law, $F\propto q_1{q}_2$

For $\text{CASE-I}:F\propto q_1{q}_2.......\left(1\right)$

For $\text{CASE-II}:F^{\prime }\propto \left(\frac{110}{100}{q}_1\right)\left(\frac{80}{100}{q}_2\right).......\left(2\right)$

From $\left(1\right)$ and $\left(2\right)$ we get,

$\frac{F^{\prime }}{F}=1.1\times 0.8\Rightarrow F^{\prime }=0.88F=0.88\times 200=176\text{N}$

Hence, option (c) is the correct answer.