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?
To determine the magnitude and direction of the electric field in the given region, we can use the formula:
Electric field (E) = Potential difference (V) / Distance (d)
In this case, the potential difference (V) is the difference in potential between the two equipotential surfaces, which is 120 volts - 75 volts = 45 volts.
The distance (d) between the two equipotential surfaces is given as 12 cm = 0.12 m.
So, substituting these values into the formula, we get:
E = 45 volts / 0.12 m = 375 volts/m
Therefore, the magnitude of the electric field in the region is 375 volts/m.
Now, to determine the direction of the electric field, we need to understand the concept of equipotential surfaces. Equipotential surfaces are imaginary surfaces in the electric field where the electric potential is constant. In this case, since there is a difference in potential between the two surfaces, the electric field must point from the higher potential (120 volts) to the lower potential (75 volts).
Hence, the direction of the electric field in the region is from the surface with 120 volts towards the surface with 75 volts.
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.