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3 point charges lie along a straight line with positive charge at the centre and negetive at the ends all are ‘a’ distance apart from each other. The energy needed to exchange the position of the central positive charge with one of the negetive charge is...............???


4 years ago

Raman Mishra
67 Points
							Let the charges be -q, +q and -q.Distance between two charges = a Energy of the system $U =\frac{ 1q_{i}q_{j}}{4\pi \varepsilon r_{ij}} ; i=1,2,3; j=1,2,3 ;i\neq j$putting the values we get Uinitial = $-\frac{3q^{2}}{8\pi \varepsilon a}$Final energy of the system, Ufinal = $-\frac{q^{2}}{8\pi \varepsilon a}$Therefore, work required = Ufinal -  Ufinal                                     = $-\frac{q^{2}}{4\pi \varepsilon a}$   However, the system will become unstable and eventually collapse.

4 years ago
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### Course Features

• 110 Video Lectures
• Revision Notes
• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions