Question:

A charge Q = 2.0 microcoulombs is placed at point A in an electric field. Point B is located 4.0 meters to the right of point A. The electric potential at point A is V(A) = 20 volts. Determine the electric potential at point B if it is 6.0 volts higher than the potential at point A.

Answer:

Given: Charge, Q = 2.0 μC Electric potential at point A, V(A) = 20 V Distance between points A and B, d = 4.0 m Increase in potential from point A to B, ΔV = 6.0 V

To determine the electric potential at point B, we can use the equation:

ΔV = -Ed

where ΔV is the potential difference, E is the magnitude of the electric field, and d is the distance between the two points.

First, we need to find the electric field strength (E) between points A and B. We know that the electric field (E) at any point due to a point charge (Q) is given by the equation:

E = k * (|Q| / r²)

where k is Coulomb's constant (k = 9.0 x 10^9 Nm²/C²), |Q| is the magnitude of the charge, and r is the distance from the charge.

In this case, the electric field at point A (E(A)) is given by:

E(A) = k * (|Q| / d²)

Substituting the given values:

E(A) = (9.0 x 10^9 Nm²/C²) * (2.0 x 10^-6 C) / (4.0 m)²

Simplifying:

E(A) = (9.0 x 10^9 Nm²/C²) * (2.0 x 10^-6 C) / 16 m²

E(A) = 1.125 x 10^5 N/C

Now, we can determine the electric potential at point B (V(B)). Since the electric field is constant between points A and B, the potential difference is directly proportional to the distance:

ΔV = V(B) - V(A) = -Ed

Rearranging the equation:

V(B) - V(A) = -E * d

Substituting the given values:

V(B) - 20 V = -(1.125 x 10^5 N/C) * 4.0 m

Simplifying:

V(B) - 20 V = -4.5 x 10^5 Nm/C

V(B) = 20 V - (-4.5 x 10^5 Nm/C)

V(B) = 20 V + 4.5 x 10^5 Nm/C

V(B) = 4.5 x 10^5 Nm/C + 20 V

Therefore, the electric potential at point B is 4.5 x 10^5 Nm/C + 20 V.