Q129 / 180 · COMEDK 2025 (25 May Shift) · Physics
Two point charges $+8 \times 10^{-18}$ and $+2 \times 10^{-18} C$ are fixed 10 cm apart. Where should we place a third charge $10^{-7} \mathrm{C}$, so that it is in equilibrium?
- A6.67 m from the charge $+2 \times 10^{-18} \mathrm{C}$
- B6.67 cm from the charge $+2 \times 10^{-18} \mathrm{C}$
- Correct answerC6.67 cm from the charge $+8 \times 10^{-18} C$
- D6.67 m from the charge $+8 \times 10^{-18} \mathrm{C}$
Solution
Since both charges are positive, the third charge must lie between them. Let $x$ be the distance of the third charge from $q_1 = +8 \times 10^{-18}$ C, so its distance from $q_2 = +2 \times 10^{-18}$ C is $(10 - x)$ cm.
For equilibrium, forces from both charges must be equal:
$\dfrac{|q_1|}{x^2} = \dfrac{|q_2|}{(10-x)^2}$
$\dfrac{8 \times 10^{-18}}{x^2} = \dfrac{2 \times 10^{-18}}{(10-x)^2}$
$\dfrac{4}{x^2} = \dfrac{1}{(10-x)^2}$
Taking square root: $\dfrac{2}{x} = \dfrac{1}{10-x}$
$2(10-x) = x \Rightarrow 3x = 20 \Rightarrow x \approx 6.67$ cm
The third charge must be placed 6.67 cm from $+8 \times 10^{-18}$ C.
For equilibrium, forces from both charges must be equal:
$\dfrac{|q_1|}{x^2} = \dfrac{|q_2|}{(10-x)^2}$
$\dfrac{8 \times 10^{-18}}{x^2} = \dfrac{2 \times 10^{-18}}{(10-x)^2}$
$\dfrac{4}{x^2} = \dfrac{1}{(10-x)^2}$
Taking square root: $\dfrac{2}{x} = \dfrac{1}{10-x}$
$2(10-x) = x \Rightarrow 3x = 20 \Rightarrow x \approx 6.67$ cm
The third charge must be placed 6.67 cm from $+8 \times 10^{-18}$ C.
Hence, the correct answer is option: 'C'.