Question

Two charges each of equal magnitude $ 3.2\times {{10}^{-19}} $ coulomb but of opposite sign form an electric dipole. The distance between the two charges is 2.4 A. If the dipole is placed in an electric field of $ 5\times {{10}^{5}} $ volt/metre, then in equilibrium its potential energy will be :

• A. $ 3\times {{15}^{-23}} $ joule
• B. $ -3.84\times {{10}^{-23}} $ joule
• C. $ -6\times {{W}^{23}} $ joule
• D. $ -2\times {{10}^{-}}^{26} $ joule

Answer 1

Answer: (B)

For an electric dipole to be in stable equilibrium, we have Potential energy $ U=-pE $ $ =-2qlE $ $ =-q\times 2l\times E $ Here: $ q=3.2\times {{10}^{-19}} $ coulomb, $ 2l=2.4{\AA} $ $ =2.4\times {{10}^{-10}}m $ $ E=5\times {{10}^{5}}volt/metre $ $ \therefore $ $ U=-3.2\times {{10}^{-19}}\times 2.4\times {{10}^{-10}}\times 5\times {{10}^{5}}J $ $ =-3.84\times {{10}^{-23}}J $