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A negative point charge -2q and a posit...

A negative point charge ` -2q` and a positive charge `q` are fixed ar a distance `l` apart. Where should a postive test charge `Q` be placed on the line connecting the charge for it to be in equlibrium ? What is the nature of the equilibrium with respect to longitudinal motions ?

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To determine where to place a positive test charge \( Q \) between a negative point charge \( -2q \) and a positive charge \( q \) fixed at a distance \( l \) apart, we need to analyze the forces acting on the test charge \( Q \). ### Step-by-Step Solution: 1. **Understanding the Setup**: - We have two charges: \( -2q \) (negative) and \( q \) (positive) separated by a distance \( l \). - We want to find a point \( x \) along the line connecting these charges where a positive test charge \( Q \) can be placed such that it experiences no net force (equilibrium). 2. **Identifying the Regions**: - The test charge \( Q \) can be placed either to the left of \( -2q \), between \( -2q \) and \( q \), or to the right of \( q \). - However, placing \( Q \) to the left of \( -2q \) would result in it being attracted to \( -2q \) and repelled by \( q \), making equilibrium impossible in that region. 3. **Analyzing the Region Between the Charges**: - We will consider placing \( Q \) between \( -2q \) and \( q \). Let’s denote the distance from \( -2q \) to \( Q \) as \( x \). Therefore, the distance from \( Q \) to \( q \) would be \( l - x \). 4. **Setting Up the Force Equations**: - The force \( F_1 \) on \( Q \) due to charge \( -2q \) is given by: \[ F_1 = k \frac{Q \cdot 2q}{x^2} \] - The force \( F_2 \) on \( Q \) due to charge \( q \) is given by: \[ F_2 = k \frac{Q \cdot q}{(l - x)^2} \] - For equilibrium, these forces must be equal in magnitude: \[ F_1 = F_2 \implies k \frac{Q \cdot 2q}{x^2} = k \frac{Q \cdot q}{(l - x)^2} \] 5. **Simplifying the Equation**: - Cancel \( k \) and \( Q \) from both sides (assuming \( Q \neq 0 \)): \[ \frac{2q}{x^2} = \frac{q}{(l - x)^2} \] - This simplifies to: \[ 2(l - x)^2 = x^2 \] 6. **Expanding and Rearranging**: - Expanding the equation: \[ 2(l^2 - 2lx + x^2) = x^2 \] - Rearranging gives: \[ 2l^2 - 4lx + 2x^2 = x^2 \implies x^2 - 4lx + 2l^2 = 0 \] 7. **Using the Quadratic Formula**: - Applying the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \): \[ x = \frac{4l \pm \sqrt{(-4l)^2 - 4 \cdot 1 \cdot 2l^2}}{2 \cdot 1} \] - This simplifies to: \[ x = \frac{4l \pm \sqrt{16l^2 - 8l^2}}{2} = \frac{4l \pm \sqrt{8l^2}}{2} = \frac{4l \pm 2\sqrt{2}l}{2} = 2l \pm \sqrt{2}l \] 8. **Finding the Valid Solution**: - The two possible solutions are: \[ x = (2 + \sqrt{2})l \quad \text{and} \quad x = (2 - \sqrt{2})l \] - Since \( x \) must be less than \( l \) (it is between the two charges), we take: \[ x = l \frac{2 - \sqrt{2}}{1} \] 9. **Nature of Equilibrium**: - To determine the nature of the equilibrium, we analyze the forces around the equilibrium position. - If \( Q \) is displaced slightly to the left, the force due to \( -2q \) will be greater than the force due to \( q \), pulling \( Q \) back to the equilibrium position (restoring force). - Conversely, if \( Q \) is displaced to the right, the force due to \( q \) will be greater, also pulling \( Q \) back to equilibrium. - Thus, the equilibrium is stable. ### Final Answer: The positive test charge \( Q \) should be placed at a distance \( \frac{l(2 - \sqrt{2})}{1} \) from the charge \( -2q \) for it to be in equilibrium. The nature of this equilibrium is stable.

To determine where to place a positive test charge \( Q \) between a negative point charge \( -2q \) and a positive charge \( q \) fixed at a distance \( l \) apart, we need to analyze the forces acting on the test charge \( Q \). ### Step-by-Step Solution: 1. **Understanding the Setup**: - We have two charges: \( -2q \) (negative) and \( q \) (positive) separated by a distance \( l \). - We want to find a point \( x \) along the line connecting these charges where a positive test charge \( Q \) can be placed such that it experiences no net force (equilibrium). ...
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