Two charges, q1 and q2, are situated on a horizontal line.  The charge on q1 is (-5q) and it is at the origin.  The charge on q2 is (+2q) and it is a distance to the right of q1.  Find the a point where the electric field is zero.

Since q1 has a charge of (-5q) and q2 has a charge of (+2q), does this mean that there is an electric field vector going from q2 with a magnitude of -3??   To solve this problem, could I first solve for the electrostatic force using Coulombs law; set the the resulting equation equal to zero, finally solving for the distance between the two charges??"

A:    The point in which the total electric field is zero is the point in which a "positive test charge q-zero" would feel no net force. You don't need to write equation for forces.   You write the formulas for the two fields created by charge q1 and charge q2, respectively, and then add these two fields and set the sum equal zero.  As you know the formula for the electric field is E = (the electric constant k_e) multiplied by the charge that creates the field and divided by the square of the distance from the charge to the point in which the field is being calculated.  So you will have E₁ = (k_e) (q1) / (r1²) and E₂ = (k_e) (q2) / (r2²).   Now substitute (-5q) for q1 and (+2q) for q2, and now the expressions of the two fields have different signs: one is positive and one is negative.  When you take the sum of these two terms and make it equal zero, q cancels out (as being common in both terms), and the resulting equation is an equation for the distance to the point in which the field is equal zero.  Now solve this equation. 

The solution of the problem is: the point with zero field is at a distance of (2.7 d) from charge q1, or (1.7 d) from charge q2.  Obviously, this point is to the right of both charges, since a "positive test charge q-zero" placed in this point would be repelled to the right by charge q2 and attracted to the left by charge q1.  These two attraction forces are equal and opposite, and the net force is equal zero.

I don't really understand what you mean in your e-mail when you say that you would like to try "an electric field vector going from q2 with a magnitude of -3??".  I guess you mean that you would like to add the charges (-5 q) and (+2 q) to a total charge of (-3 q).  Is that what you mean?  In this case, that would be incorrect.  Leave the individual charges in the points where they are located in space.  Each of the creates a field around itself.  The two charges cannot be merged (added) together.  The real electric field in each point in space is the net sum (algebraic sum) of the individual fields created by the individual charges at the specific point where the net field is calculated.

It is very important to study the theory and understand it very well, before attempting to solve problems.  You will realize in this class that the problems are simple, and most often involve only simple algebraic calculations.  The tricky part (and the most important one) is to understand in detail the abstract concepts related to fields created by electric charges positioned in various geometric configurations in space.  Some electrical technologists can apply the laws of electricity (designing, building, or troubleshooting complex electrical circuits) without needing to understand the abstract concepts.   For scientists and engineers, though, such understanding is essential.

For in-depth study of the concept of electric field, I advise you to also use the "Interactive Java Applet: Electric Field and Potential Lines" from our "Electric Fields" and "Electric Potential" tutoring web pages located at: http://www.slcc.edu/schools/hum_sci/physics/tutor/2220/e_fields/java/index.html.

Q:     A proton and an electron form two corners of an equilateral triangle with sides of length
3 x 10 ^-6 m.  What is the magnitude of their net electric field at the third corner?

A:    For this problem we draw as a diagram of the charges and find E_net and point P(x,y).

Known: E = ke q/r2 q1 = p+, q2 = e-,   q1 = q2 = 1.6 x 10 -19 C r = a = 3 x 10-6 m E2y = -E1y Solve for Enet : (Enet) = (Enet)x + (Enet)y (Enet)y   = E1y + E2y =  E1y - E1y = 0 (Enet)x   = E1x + E2x              = E1x cos + E2x cos              = E1x cos 60º + E2x cos 60º  =  (E1x + E2x) cos 60º              = ( keq1/a2 +  keq2/a2) cos 60º              = 4.8 x 10-4 V/m (Enet) = (Enet)x + (Enet)y = (Enet)x = 4.8 x 10-4 V/m

Q:     How does the electric field vary with distance from the center of a sphere that is uniformly charged?

A:     If the sphere is uniformly charged, it means that it is an insulator.  Inside the spherical insulator, the electric field is proportional to the distance from the center, and outside the sphere the electric field is inversely proportional to the square of the distance.