Two equipotential surfaces surround a +1.63 10-8-C point charge. How far is the 204-V surface from the 81.5-V surface?
Answer in m.
To find the distance between two equipotential surfaces, we need to take the following steps:
Step 1: Recall that the potential difference, also known as voltage, between two equipotential surfaces is given by the formula:
ΔV = V2 - V1
Where ΔV is the potential difference, V2 is the voltage of the second surface, and V1 is the voltage of the first surface.
In this case, we are given the voltage of two surfaces: 204 V and 81.5 V.
Step 2: Calculate the potential difference:
ΔV = 204 V - 81.5 V
ΔV = 122.5 V
Step 3: Recall that the potential difference is directly proportional to the distance between the equipotential surfaces. This means that as the potential difference increases, so does the distance between the surfaces.
Step 4: For a point charge, the potential difference at a distance r from the charge is given by the formula:
ΔV = k(Q/r)
Where ΔV is the potential difference, k is the electrostatic constant (9 × 10^9 N m^2/C^2), Q is the charge in coulombs, and r is the distance from the charge in meters.
Step 5: Rearrange the formula to solve for r:
r = k(Q/ΔV)
Step 6: Substitute the known values into the formula:
r = (9 × 10^9 N m^2/C^2) * (+1.63 × 10^(-8) C) / (122.5 V)
Step 7: Calculate the distance:
r ≈ 1.206 × 10^(-3) m
Therefore, the distance between the 204 V and 81.5 V equipotential surfaces is approximately 1.206 × 10^(-3) meters.