15. The test charge creates its own electric field. The measured electric field is the sum of the original electric field plus the field of the test charge. By making the test charge small, the field that it causes is small. Therefore the actual measured electric field is not much different than the original field.  Also, if the test charges are large, their fields might significantly re-distribute the charges causing the original field, and then the measurement would not represent the field of the original configuration of charges.

16. When determining an electric field, it is best, but not required, to use a positive test charge. A negative test charge would be fine for determining the magnitude of the field. But the direction of the electrostatic force on a negative test charge will be opposite to the direction of the electric field. The electrostatic force on a positive test charge will be in the same direction as the electric field. In order to avoid confusion, it is better to use a positive test charge. If we used a negative test charge but wanted to have the same result for the electric field, we would have to define E = -F/q, q < 0.

17. See Fig. 21–35(b). A diagram of the electric field lines around two negative charges would be just like this diagram except that the arrows on the field lines would point towards the charges instead of away from them. The distance between the charges is l.

18. The electric field will be strongest to the right of the positive charge (between the two charges) and weakest to the left of the positive charge. To the right of the positive charge, the contributions to the field from the two charges point in exactly the same direction, and therefore add. To the left of the positive charge, the contributions to the field from the two charges point in exactly opposite directions, and therefore subtract. Note that this should be confirmed by the density of field lines in Fig. 21–35a.

19. At point C, the positive test charge would experience zero net force.  At points A and B, the direction of the force on the positive test charge would be the same as the direction of the field. This direction is indicated by the arrows on the field lines. The strongest field is at point A, followed (in order of decreasing field strength) by B and then C. The field strength can be inferred by the density of the field lines at each location.

20. Electric field lines can never cross because they give the direction of the electrostatic force on a positive test charge. If they were to cross, then the force on a test charge at a given location would be in more than one direction. This is not possible, because the electric force has a unique direction at each point in space.

21. From rule 1: A test charge would be either attracted directly towards or repelled directly away from a point charge, depending on the sign of the point charge. So the field lines must be directed either radially towards or radially away from the point charge.
From rule 2: The magnitude of the field due to the point charge only depends on the distance from the point charge. Thus, the density of the field lines must be the same at any location around the point charge, for a given distance from the point charge.
From rule 3: If the point charge is positive, the field lines will originate from the location of the point charge. If the point charge is negative, the field lines will end at the location of the point charge.
Based on rules 1 and 2, the lines are radial and their density is constant for a given distance. This is equivalent to saying that the lines must be symmetrically spaced around the point charge.

22. The two charges are located as shown in the diagram.
(a) If the signs of the charges are opposite then the point on the line where E = 0 will lie to the left of Q. In that region the electric fields from the two charges will point in opposite directions, and the point will be closer to the smaller charge.
(b) If the two charges have the same sign, then the point on the line where E = 0 will lie between the two charges, closer to the smaller charge. In this region, the electric fields from the two charges will point in opposite directions.

23. The electric field at point P would point in the negative x-direction. The magnitude of the field would be the same as that calculated for a positive distribution of charge on the ring:

E=(1/4*pi*e_0) * [ (|Q|x)/(x^2+a^2)^3/2 ] 

24. We assume that there are no other forces (like gravity) acting on the test charge. The direction of the electric field line shows the direction of the force on the test charge. The acceleration is always parallel to the force by Newton’s second law, so the acceleration lies along the field line. If the particle is at rest initially and then released, the initial velocity will also point along the field line, and the particle will start to move along the field line. However, once the particle has a velocity, it will not follow the field line unless the line is straight. The field line gives the direction of the acceleration, or the direction of the change in velocity.

25. The value measured will be slightly less than the electric field value at that point before the test charge was introduced. The test charge will repel charges on the surface of the conductor and these charges will move along the surface to increase their distances from the test charge. Since they will then be at greater distances from the point being tested, they will contribute a smaller amount to the field.

26. The motion of the electron in Example 21–17 is analogous to projectile motion. In the case of the gravitational force, the acceleration of the projectile is in the same direction as the field and has a value of g no matter what the projectile is. In the case of an electron in an electric field, the direction of the acceleration of the electron and the field direction are opposite, and the value of the acceleration depends on the mass of the electron, the charge of the electron, and the magnitude of the electric field.

27. If an electric dipole is placed in a nonuniform electric field, the charges of the dipole will experience forces of different magnitudes whose directions also may not be exactly opposite. The sum of these forces might not be zero, and so there can be a net force on the dipole.

28. (a) Initially, the dipole will rotate clockwise about an axis through point 0 (its center of mass), out of the page. It will “overshoot” the equilibrium position (parallel to the field lines), come momentarily to rest and then spin counterclockwise. The dipole will continue to oscillate back and forth if no damping forces are present. If there are damping forces, the amplitude will decrease with each oscillation until the dipole comes to rest aligned with the field.
(b) There will be a larger force on the negative charge, since the electric field is stronger at the location of the negative charge. Thus the dipole will accelerate to the left as it rotates. Once again, the dipole will “overshoot” the horizontal equilibrium orientation. When that happens, the positive charge will have a larger force than the negative charge, causing an acceleration to the right. The dipole will slow it’s leftward motion, and eventually start to move back to the right, even as the torque causes counterclockwise acceleration and the rotation also shifts to counterclockwise. Thus, it will oscillate both horizontally and rotationally.