Question

Suppose there is a uniform electric field pointing in the positive x-direction with a magnitude of 5.0 V/m. The electric potential is measured to be 50 V at the position x = 10 m. What is the electric potential at other positions?

Position [m] = (−20)--- (0.00) ---(10)--- (11)--- (99)
Electric Potential [V]=

Answer 1

The electric potential at various positions in a uniform electric field can be determined using the relationship ΔV = -E • (xf - xi). For a given field of 5.0 V/m and an initial potential of 50 V at x = 10 m, the electric potentials at x = -20 m, 0 m, 10 m, 11 m, and 99 m are 200 V, 100 V, 50 V, 45 V, and -395 V respectively.

Step-by-step explanation:

To calculate the electric potential at various positions in a uniform electric field, you can use the relationship ΔV = -E • (xf - xi), where ΔV is the change in potential, E is the magnitude of the electric field, and xf and xi are the final and initial positions respectively. In this situation, the electric field E is 5.0 V/m. If we know the potential at x = 10 m is 50 V, we can determine the potential at any other position x by using the given electric field.

For example, the potential at x = 0 m would be 50 V + (5.0 V/m)(10 m - 0 m) = 100 V. Similarly, we can find the potential at x = -20 m, x = 11 m, and x = 99 m, by calculating the change in potential for each respective position from the known position at x = 10 m.

Specifically:

• At x = -20 m: V = 50 V + (5 V/m)(10 m - (-20 m)) = 200 V
• At x = 0.00 m: V = 50 V + (5 V/m)(10 m - 0 m) = 100 V
• At x = 10 m: V = 50 V (given)
• At x = 11 m: V = 50 V + (5 V/m)(10 m - 11 m) = 45 V
• At x = 99 m: V = 50 V + (5 V/m)(10 m - 99 m) = -395 V

To maintain clarity, it's important to keep track of the sign when subtracting distances and to apply the negative sign for the change in potential when the field is in the opposite direction of the displacement.

Answer 2

Electric potential at position, x = -20 m, = -100 V

Electric potential at position, x = 0 m, = 0

Electric potential at position, x = 10 m, = 50 V

Electric potential at position, x = 11 m, = 55 V

Electric potential at position, x = 99 m, 495 V

Step-by-step explanation:

Given;

magnitude of electric field, E = 5.0 V/m

at position x = 10 m, electric potential = 50 V

Electric potential at position, x = -20 m

V = Ex

V = 5 (-20)

V = -100 V

Electric potential at position, x = 0 m

V = Ex

V = 5(0)

V = 0

Electric potential at position, x = 10 m

V = Ex

V = 5(10)

V = 50 V

Electric potential at position, x = 11 m

V = Ex

V = 5(11)

V = 55 V

Electric potential at position, x = 99 m

V = Ex

V = 5(99)

V = 495 V