The poll question stem, which was posted throughout the week, is shown to the right. Here's my answer.
Let's start with direction. The electric field produced by a positive charge points away from the charge; the electric field produced by a negative charge points toward the charge. Since a negative charge produces this electric field, the direction is "toward the charge."
Now for the magnitude of the electric field. First of all, our students must be taught to use the variables given in the problem. I know that the textbook says that the electric field due to a charge is E = kq/r^2. This problem defined the distance from the charge as "L". Thus, we must use "L" and not "r" in the answer.
Secondly, our students must understand what the word "magnitude" means. All vector quantities, including electric field, include an amount of something (the magnitude) and a direction. For example, a velocity might be 30 m/s toward the south. In this case, "30 m/s" is the velocity's magnitude, and "south" is the direction.
Now, when a vector is constrained to one direction (or when we're dealing with components of a 2-d or 3-d vector), it is mathematically convenient to define a negative and a positive direction. If I'm working a kinematics problem with this car, I might call north the positive direction, in which case the velocity vector can be plugged in simply as "-30 m/s". But, importantly, the magnitude of this vector is still "30 m/s." The magnitude of a vector can not be negative!
Electric field is a vector. When using equations regarding electric fields, never plug in the sign of the charge! use the equation to find the magnitude of an electric field or force. Then, use memorized facts to determine the direction of an electric field or force.
In this case, then, the negativeness of the charge producing the electric field does not affect the field's magnitude. The field has magnitude kq/L^2. The direction of the electric field is toward the charge.