The potential energy at point D at x = 8 m is -2000 V and at point B at x = 2 m is +400 V. What is the magnitude and direction of the electric field?
To find the magnitude and direction of the electric field, we can use the equation:
V = E * d,
where V is the potential difference, E is the electric field strength, and d is the distance.
Given:
Potential at point D, V(D) = -2000 V
Potential at point B, V(B) = +400 V
Distance between points D and B, d = 8 m - 2 m = 6 m
Now, let's calculate the electric field magnitude:
Magnitude of potential difference (delta V) = |V(D) - V(B)|
= |-2000 V - 400 V|
= 2400 V
Electric field magnitude (E) = delta V / d
= 2400 V / 6 m
= 400 V/m
Therefore, the magnitude of the electric field is 400 V/m.
To determine the direction of the electric field, we need to consider the potential difference. In this case, since the potential at point D (V(D)) is negative, and the potential at point B (V(B)) is positive, we can conclude that the electric field is pointing from point D to point B.
Hence, the direction of the electric field is from point D to point B.