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⚛️ Physics  ·  Electric Charges and Fields  ·  NEET & JEE

Three charges, +q, +q and -q, are placed at the vertices of an equilateral triangle of side 0.2 m. If q = 2 µC, the total electrostatic potential energy of the system is:

Answer: -0.18 J.

  • A +0.09 J
  • B +0.18 J
  • C -0.36 J
  • D -0.18 J

Correct answer: D. -0.18 J

Explanation: The pair energies are (+q)(+q) = +kq<sup>2</sup>/a and the two (+q)(-q) pairs = -kq<sup>2</sup>/a each. Sum U = -kq<sup>2</sup>/a = -(9 &times; 10<sup>9</sup> &times; 4 &times; 10<sup>-12</sup>)/0.2 = -0.18 J.

Field Lines: Single Charge vs Electric Dipole+isolated +charge: lines radiate outward, all directions+dipole: lines curve from + to − (denser between them)

Field lines always point from positive to negative charge, are denser where the field is stronger, and never cross; an isolated positive charge has lines radiating symmetrically outward, while a dipole's lines curve from the positive to the negative charge.

Concept context

Coulomb's law, electric field, potential, capacitors, Gauss's law.

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