An electron enters a region of space where the potential difference between two equipotential surfaces are 12 cm apart. One surface has a potential of 120 volts and the other potential of 75 volts. What is the magnitude and direction of the electric field in the region?

Answer 1

Memorize this:

energy= Volts*charge=E*distance

From that
E=Volts*g/distance

direction is the direction a + test charge will go (toward the negative), away from the most positive.