The Setup
A Forest of Infinite Sheets
Imagine you are standing in a vast, empty space, and suddenly, six infinitely large, incredibly thin sheets appear before you. They are perfectly parallel, spaced exactly 1 μm apart. This isn't just a random arrangement; these sheets are non-conducting and carry uniform surface charge densities. We are presented with two such configurations, Configuration I and Configuration II, and our mission is to decode the electric fields and potential differences hidden within them.
The Master Tool
Superposition Principle
Before we dive into the complex configurations, let's recall our fundamental tool. What is the electric field generated by a single, infinite, non-conducting sheet? It's a beautifully simple expression:
This field is uniform, meaning it doesn't weaken as you move away from the sheet. It points away from positive charges and towards negative charges. But what happens when we have multiple sheets? Enter the Superposition Principle. The net electric field in any region is simply the vector sum of the electric fields produced by each individual sheet. To keep things organized, let's establish a sign convention: fields pointing to the right are positive, and fields pointing to the left are negative.
Analyzing Configuration I
The Dance of Charges
Let's focus on Configuration I. The charge densities on the six sheets are +σ0,−σ0,+σ0,−σ0,+σ0,−σ0. We need to evaluate the electric field in Region 4, which lies between the 4th and 5th sheets.
To find the net field here, we sum the contributions from all sheets. The sheets to the left of Region 4 (sheets 1, 2, 3, and 4) create fields pointing to the right (if positive) or left (if negative). The sheets to the right (sheets 5 and 6) create fields pointing in the opposite directions.
E4=2ϵ01[(σ1+σ2+σ3+σ4)−(σ5+σ6)]
Substituting the given charge densities:
E4=2ϵ01[(σ0−σ0+σ0−σ0)−(σ0−σ0)]
Notice the elegant cancellation! The sum of charges on the left is zero, and the sum on the right is also zero. Therefore, the net electric field in Region 4 is exactly zero. Statement (A) is absolutely correct.
Now, let's calculate the potential difference between the first and last sheets. We need to find the electric field in every region and multiply it by the separation distance d.
- E1=2ϵ01[σ0−(−σ0+σ0−σ0+σ0−σ0)]=ϵ0σ0
- E2=2ϵ01[(σ0−σ0)−(σ0−σ0+σ0−σ0)]=0
- E3=2ϵ01[(σ0−σ0+σ0)−(−σ0+σ0−σ0)]=ϵ0σ0
- E4=0 (as calculated earlier)
- E5=2ϵ01[(σ0−σ0+σ0−σ0+σ0)−(−σ0)]=ϵ0σ0
The total potential difference is the sum of the potential drops across each region:
ΔVI=(E1+E2+E3+E4+E5)d=(ϵ0σ0+0+ϵ0σ0+0+ϵ0σ0)d=ϵ03σ0d
Plugging in the given values (σ0=9×10−6 C/m2, d=10−6 m, ϵ0=9×10−12 F/m):
ΔVI=9×10−123×9×10−6×10−6=3 V
Statement (C) claims it is 5 V, so it is incorrect.
Analyzing Configuration II
Fractional Charges
Now, let's turn our attention to Configuration II. The charge densities are +2σ0,−σ0,+σ0,−σ0,+σ0,−2σ0. We need to check the electric field in Region 3.
Applying our superposition formula again:
E3=2ϵ01[(2σ0−σ0+σ0)−(−σ0+σ0−2σ0)]
E3=2ϵ01[2σ0−(−2σ0)]=2ϵ01[σ0]=2ϵ0σ0
Statement (B) claims the field is ϵ0σ0, which is incorrect.
Finally, let's find the potential difference for Configuration II. We calculate the field in each region:
- E1=2ϵ01[2σ0−(−σ0+σ0−σ0+σ0−2σ0)]=2ϵ0σ0
- E2=2ϵ01[(2σ0−σ0)−(σ0−σ0+σ0−2σ0)]=−2ϵ0σ0
- E3=2ϵ0σ0
- E4=−2ϵ0σ0
- E5=2ϵ0σ0
The total potential difference is:
ΔVII=(E1+E2+E3+E4+E5)d=(2ϵ0σ0−2ϵ0σ0+2ϵ0σ0−2ϵ0σ0+2ϵ0σ0)d=2ϵ0σ0d
ΔVII=2×9×10−129×10−6×10−6=0.5 V
Statement (D) claims the potential difference is zero, which is also incorrect.
The Final Verdict
After a rigorous mathematical investigation, we have systematically dismantled the incorrect options. The only statement that stands true to the laws of electrostatics is (A). This problem beautifully illustrates the power of the superposition principle and the importance of keeping track of signs when dealing with vector fields.
