Question

The electric potential along the x-axis is V=100e−2xV, where x is in meters.

A) What is Ex at x=1.0m?
B) What is Ex at x=1.6m? Express your answer to two significant figures and include the appropriate units.

Answer 1

The electric field Ex at x=1.0m is calculated using the negative gradient of the potential, yielding 200e−2 N/C. For x=1.6m, the same method gives a field strength of approximately 9.1 N/C to two significant figures.

Step-by-step explanation:

The student is asking how to calculate the electric field (Ex) at two different positions along the x-axis from a given electric potential equation. The electric potential is given by V=100e−2x V where x is the distance in meters. To find the electric field strength, we will need to calculate the negative gradient of the potential.

For x=1.0m:

Ex = -dV/dx = -(-2)(100e−2(1.0)) = 2(100e−2) = 200e−2 N/C

For x=1.6m:

Ex = -dV/dx = -(-2)(100e−2(1.6)) = 2(100e−2.6) = 200e−2.6 N/C. After evaluating, this approximates to Ex ≈ 9.1 N/C to two significant figures.