9. [03.03] When comparing two circuits, you note that circuit one has twice the voltage and half the resistance of circuit two.

That's a very interesting observation.  When you notice that, does it stimulate any questions in your mind due to your natural curiosity ?

For example, do you walk away wondering, perhaps, how the current in the two circuits might compare ?  If so, you've come to the right place.  Read on:

Take the form of Ohm's Law for current . . . I = V / R . . . and apply it to both circuits:

For the second circuit:  I₂ = V / R

For the first circuit:  

I₁ = (2V) / (R/2)

Multiply numerator and denominator by 2 :

I₁ = (4V) / R

I₁ = 4 (V / R)

I₁ = 4 I₂  

two charges are fixed in place on the x-axis as shown. the charge at x = 0 is negative and has magnitude q = 2 μcoulombs (1 μcoulomb = 10-6 coulombs). the charge at x = d, (d = +12 cm), is positive and has magnitude q = 4 μcoulombs. find a point x = xo along the x-axis at which the electric field is zero.