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Q.A charge 'q' is placed at the midpoint of the line joining two equal charges 'Q'. For what value of 'q' will the entire charge system remain in equilibrium? OR Derive the expression for the torque acting on an electric dipole placed in a uniform electric field.

Tripura TbseHigher Secondary (+2 Stage) Examination 2024Subjective· 2mImportance★★★★★
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The middle charge is automatically in equilibrium by symmetry for ANY value of q; the real condition comes from requiring the two end charges Q to also be in equilibrium, which forces q = -Q/4.

Let the two equal charges +Q+Q be placed at points A (x=0x=0) and B (x=2ax=2a), with charge qq at the midpoint M (x=ax=a).

Equilibrium of q: By symmetry, the two forces on qq due to the equal charges QQ at A and B (each at distance aa) are equal in magnitude and act in opposite directions along the line — so qq is automatically in equilibrium at the midpoint, for ANY value or sign of qq. This alone does not fix qq.

Equilibrium of the whole system: For the SYSTEM to be in equilibrium, the charges at A and B must ALSO be in equilibrium (not just qq). Consider the charge QQ at A. It experiences:

  • A repulsive force from the other charge QQ at B (separation 2a2a): F1=kQ2(2a)2=kQ24a2F_1 = \dfrac{kQ^2}{(2a)^2} = \dfrac{kQ^2}{4a^2}, directed away from B (pushing A further left).
  • A force from charge qq at M (separation aa): F2=kQqa2F_2 = \dfrac{kQq}{a^2}. …

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