Three charges are at the corners of an isosceles tri-angle as shown in Fig. E21.57. The charges form a dipole. (a) Find the force (magnitude and direction) the - 10.00-µC charge exerts on the dipole. (b) For an axis perpendicular to the line connecting the charges at the mid-point of this line, find the torque (magnitude and direction) exerted on the dipole by the - 10.00-µC charge.
Solution 61E Introduction We have to calculate the force and torque on the dielectric by the 10 C charge. Step 1: Above figure represents the force on each charge of the dipole. The magnitude of the force on each charge will be same as the distance between the charges and the magnitude of the charges are same. But as shown in the figure one force will be attractive and other will repulsive. The direction of the forces are shown in the figure. The magnitude of the force is |1 |2q | |F 1 = |F 2 = |F| = k r2 Here k = 1 = 8.99 × 10 Nm C 2 2 40 2 6 We also have r = 2.00 cm = 2.00 × 10 cm, |q | 1 5.00 C = 5.00 × 10 C , and 6 |q2| = 10 × 10 C , Hence the magnitude of the force is 6 6 |F| = (8.99 × 10 Nm C ) 2 2 (5.00×10 C)(2×12 C) = 1.124 × 10 N3 (2.00×10 m) Now the x and y component of the force is given by |F x = |F 1x= |F |2x F cos And |F | = |F | = |F | = F sin y 1y 2y Now we have (3.00 cm)/2 sin = 2.00 cm = 0.75 cos = 1 in = 0.66 Hence the x component is |F x = |F 1x= |F |2x (1.124 N)(0.66) = 742 N and |F y = |F 1y= |F | 2y(1.124 N)(0.75) = 843 N From the figure we can see that the x components are opposite to each other, hence the total x component will be zero. But the y components are parallel hence the total component will be 3 |Fyt = 2|F y = 2(843 N) = 1.69 × 10 N 3 Hence the total force on the dipole is 1.69 × 10 N and it is downward.