A dipole considts of two opposite charges q and -q separated by a fixed distance

Question

A dipole considts of two opposite charges q and -q separated by a fixed distance d. The dipose is placed in an electric field in the + x direction of magnitude E. The dipole axis makes an angle above the x axis. Calculate the magnitude of the net ele3ctric force acting on the dipose. Calculate the net torque acting on the dipose due to the electric forces as a function of the angle. Counterclockwise torque is positive. Evaluate the net torque for angle = 0, angle = 30.6, angle = 90. Let q = 2.9 uC, d = 7.0 cm, and E = 2.0 x 10^4 N;C. torque is (N.m)

Explanation / Answer

Given data: Let the two charges are q and -q, and magnitude of charge, q = 2.9 *10-6 C Charges are seperated by a distance, d = 0.07 m Magnitude of electric field is, E = 2.0 *104 N/C -------------------------------------------------------------------------------- Solution: Magnitude of net electric force acting on the dipole is,                         F = qE                            = (2.9 *10-6 C) (2.0 *104 N/C)                            = 5.8 *10-2 N ----------------------------------------------------------------------------------- The net torque acting on the dipole is,                          = F *2d sin                       ...... (1)                            = qE *2d sin                             = pE sin      [since p = q *2d] ------------------------------------------------------------------------------------ If = 00 Net torque is calculated using equation (1),                           = F *2d sin                                                    = (5.8 *10-2 N) *2 *(0.07 m) *sin0                              = 0 If = 30.60 Net torque is,                           = F *2d sin                                                    = (5.8 *10-2 N) *2 *(0.07 m) *sin30.6                              = 0.004 N.m If = 900 Net torque is,                           = F *2d sin                                                    = (5.8 *10-2 N) *2 *(0.07 m) *sin90                              = 0.008 N.m ----------------------------------------------------------------------------------- The net torque acting on the dipole is,                          = F *2d sin                       ...... (1)                            = qE *2d sin                             = pE sin      [since p = q *2d] ------------------------------------------------------------------------------------ If = 00 Net torque is calculated using equation (1),                           = F *2d sin                                                    = (5.8 *10-2 N) *2 *(0.07 m) *sin0                              = 0 If = 30.60 Net torque is,                           = F *2d sin                                                    = (5.8 *10-2 N) *2 *(0.07 m) *sin30.6                              = 0.004 N.m If = 900 Net torque is,                           = F *2d sin                                                    = (5.8 *10-2 N) *2 *(0.07 m) *sin90                              = 0.008 N.m Net torque is,                           = F *2d sin                                                    = (5.8 *10-2 N) *2 *(0.07 m) *sin30.6                              = 0.004 N.m If = 900 Net torque is,                           = F *2d sin                                                    = (5.8 *10-2 N) *2 *(0.07 m) *sin90                              = 0.008 N.m Net torque is,                           = F *2d sin                                                    = (5.8 *10-2 N) *2 *(0.07 m) *sin90                              = 0.008 N.m

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