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प्रश्न
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.
उत्तर
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,
\[F - F' = 0\]
\[\Rightarrow F = F'\]
\[ \Rightarrow \frac{9 \times {10}^9 \times 2 \times {10}^{- 6} \times q}{x^2} = \frac{9 \times {10}^9 \times {10}^{- 6} \times q}{\left( x - 10 \right)^2}\]
\[ \Rightarrow x^2 = 2 \left( x - 10 \right)^2 \]
\[ \Rightarrow x^2 - 40x + 200 = 0\]
\[ \Rightarrow x = 20 \pm 10\sqrt{2} \text{ m} \]
\[ \Rightarrow x = 34 . 14 \text{ cm } ( \because x \neq 20 - 10\sqrt{2})\]
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