Three point charges lie along a straight line as shown in th

Three point charges lie along a straight line as shown in the figure below, where q_1 = S.46 pC, q_2 = 1.66 muC, and q_3 = -2.14 muC. The separation distances are d_1 = 3.00 cm and d_2 = 2.00 cm. Calculate the magnitude and direction of the net electric force on each of the charges.

Solution

The force on q1 due to q2 and q3 is F1 = F(q1.q2) + F (q1.q3)

    coloumb force between q1 and q2 = F (q1.q2) = k q1 *q2 / d1^2

                                                                          = 8.98 * 10^9 * 5.46*1.66*10^-12 / 0.03 ^2

                                                                        = 90.434 newtons which is replusive in nature

coloumb force between q1 and q3 = F (q1.q3) = k q1 *q3 / d1^2

                                                                                = 8.98 * 10^9 * 5.46* -2.14*10^-12 / 0.03 ^2

                                                                              = 41.970 N attractive

net force on q1 is 90.434 - 41.970 N = 48.464 Repulsive and the direction is away from q2 and q3

by simillare method the force on q2due to q3 F(q2.q3) = 8.98 * 10^9 *1.66*-2.14*10^-12 / 0.03 ^2

                                                                            = 76.754 N towards q3

For q2

force on q2 due to q1 = force on q1 due to q2 = 90.434 repulsive away froom q1

by newtons 3rd law the action force is equal and opposite to reaction force

                      net force on q2 due to q1 and q3 is = 90.434 + 76.754 N = 167.188 N towards q3

For q3

The force of attraction on q3 due to q1 and q2 = 41.970+76.754 = 118.724 N towards q2 and q1