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, $2l=2.4\mathring{A}$ $=2.4\times {10}^{-10}m$ $E=5\times {10}^{5}volt/metre$ $\therefore$ $U=-3.2\times {10}^{-19}\times 2.4\times {10}^{-10}\times 5\times {10}^{5}J$ $=-3.84\times {10}^{-23}J$