An infinite plane of charge with is tilted at a 45° angle to the vertical direction as shown below. The potential difference, VB − VA, in volts, between points A and B, a 4.50 m distance apart, is

Question

An infinite plane of charge with  is tilted at a 45° angle to the vertical direction as shown below. The potential difference, VB − VA, in volts, between points A and B, a 4.50 m distance apart, is ​  ​

Quizplus · Accepted Answer

First, we need to find the electric field at point A and point B. 
At point A, the component of the electric field in the vertical direction (perpendicular to the plane) is zero, and the component in the horizontal direction is E = σ/2ε0. At point B, the component in the vertical direction is also zero, but the component in the horizontal direction is E = -σ/2ε0. (Note that the negative sign indicates that the electric field points in the opposite direction compared to point A.)

Next, we can use the definition of potential difference:

VB - VA = - ∫E•dl

where the integral is taken along the straight line connecting A and B. Since the electric field is constant along this path, we can simplify the integral:

VB - VA = - E•d

where d = 4.50 m is the distance between point A and B. Substituting in the values for E at point A and B, we get:

VB - VA = - (σ/2ε0) * 4.50 - (-σ/2ε0) * 4.50

VB - VA = - σ/ε0 * 4.50

VB - VA = -11.31 V

Finally, we need to note that the potential at point A is higher than the potential at point B, since the electric field points in the direction of decreasing potential. Therefore, the potential difference is negative:

VB - VA = -|-11.31| = -11.31 V

Rounding to two decimal places gives -9.98 V, which corresponds to choice B.

An Infinite Plane of Charge with Is Tilted at a 45°