Question

Suppose the second charge in the previous problem is −1.0 × 10⁻⁶ C. Locate the position where a third charge will not experience a net force. 

Answer 1

Given: 

\[q_1  = 2 \times  {10}^{- 6}   C\] 

\[ q_2  =  - 1 \times  {10}^{- 6}   C\]

Since both the charges are opposite in nature, the third charge cannot be placed between them. Let the third charge, q, be placed at a distance of x cm from charge q₁, as shown in the figure.

By Coulomb's Law, force,           

\[F = \frac{1}{4\pi \epsilon_0}\frac{Q_1 Q_2}{r^2}\]

So, force on charge q due to q₁,

\[F = \frac{9 \times {10}^9 \times 2 . 0 \times {10}^{- 6} \times q}{x^2}\]

Force on charge q due to q₂,

\[F' = \frac{9 \times {10}^9 \times {10}^{- 6} \times q}{\left( x - 10 \right)^2}\]

\[\text{ According to the question, } \]
\[F - F' = 0\]

\[\Rightarrow F = F'\] 

\[ \Rightarrow \frac{9 \times {10}^9 \times 2 \times {10}^{- 6} \times q}{x^2} = \frac{9 \times {10}^9 \times {10}^{- 6} \times q}{\left( x - 10 \right)^2}\] 

\[ \Rightarrow  x^2  = 2 \left( x - 10 \right)^2 \]

\[ \Rightarrow  x^2  - 40x + 200 = 0\] 

\[ \Rightarrow x = 20 \pm 10\sqrt{2}  \text{ m} \] 

\[ \Rightarrow x = 34 . 14  \text{ cm }    ( \because x \neq 20 - 10\sqrt{2})\]