The potential difference ΔV between two points equals what integral of the electric field?

The potential difference ΔV between two points equals what integral of the electric field?

• A. ΔV = ∫ E · dl
• B. ΔV = ∫ B · dl
• C. ΔV = − ∫ E · dl
• D. ΔV = ∇ × E

Answer: (C)

Electric potential is defined so that the electric field points in the direction of decreasing potential. This gives the relationship E = -∇V, which leads to the potential difference between two points being the negative of the line integral of the field along the path from the first to the second: ΔV = V(b) - V(a) = - ∫_a^b E · dl. The minus sign is essential because moving a positive test charge a small amount along the field reduces the potential by exactly the amount of the line integral of E, per unit charge. In electrostatics, where charges aren’t changing in time, this line integral is path-independent, so ΔV is well defined regardless of the path taken. The magnetic field integral isn’t related to the potential difference, and the curl of E relates to changing magnetic fields (Faraday’s law) rather than directly giving a potential difference.

Electric potential is defined so that the electric field points in the direction of decreasing potential. This gives the relationship E = -∇V, which leads to the potential difference between two points being the negative of the line integral of the field along the path from the first to the second: ΔV = V(b) - V(a) = - ∫_a^b E · dl. The minus sign is essential because moving a positive test charge a small amount along the field reduces the potential by exactly the amount of the line integral of E, per unit charge. In electrostatics, where charges aren’t changing in time, this line integral is path-independent, so ΔV is well defined regardless of the path taken. The magnetic field integral isn’t related to the potential difference, and the curl of E relates to changing magnetic fields (Faraday’s law) rather than directly giving a potential difference.