Consider the arrangement of parallel plates of the previous problem. If 1.0 nC charge is given to the upper plate instead of the middle, what will be the potential difference (in V) between (a) the upper and the middle plates and (b) the middle and the lower plates ? If Capacitance of each plate is 50 nF .

Answer

To solve the problem, we need to find the potential difference between the plates when a charge of 1.0 nC is given to the upper plate instead of the middle plate. The capacitance of each plate is given as 50 nF (nanoFarads).

### Step-by-Step Solution:

1. **Identify the Charges on the Plates:**
- Given that 1.0 nC (nanoCoulomb) charge is placed on the upper plate (let's call it Plate 1), we need to determine how this affects the charges on the other plates.
- The charge on Plate 1 is +1.0 nC.
- To maintain neutrality, the charge on Plate 2 (middle plate) will be -0.5 nC, and the charge on Plate 3 (lower plate) will also be +0.5 nC.

2. **Calculate the Potential Difference Between Upper and Middle Plates (V12):**
- The formula for potential difference (V) between two plates is given by:
\[ V = \frac{Q}{C} \]
- Here, \(Q\) is the charge on the upper plate (Plate 1), which is +0.5 nC (since we consider the effect of the middle plate charge), and \(C\) is the capacitance of the plates, which is 50 nF.
- Convert the values to standard units:
- \(Q = 0.5 \times 10^{-9} \, \text{C}\)
- \(C = 50 \times 10^{-9} \, \text{F}\)
- Now, substitute the values into the formula:
\[ V_{12} = \frac{0.5 \times 10^{-9}}{50 \times 10^{-9}} = \frac{0.5}{50} \, \text{V} = 0.01 \, \text{V} = 10^{-2} \, \text{V} \]

3. **Calculate the Potential Difference Between Middle and Lower Plates (V23):**
- For the potential difference between the middle plate (Plate 2) and the lower plate (Plate 3), we again use the same formula:
\[ V_{23} = \frac{Q_{2}}{C} \]
- The charge on Plate 2 is -0.5 nC, but we will consider the absolute value for the potential difference calculation.
- Thus, we have:
\[ V_{23} = \frac{0.5 \times 10^{-9}}{50 \times 10^{-9}} = \frac{0.5}{50} \, \text{V} = 0.01 \, \text{V} = 10^{-2} \, \text{V} \]

### Final Answers:
- (a) The potential difference between the upper and middle plates (V12) is \(10^{-2} \, \text{V}\).
- (b) The potential difference between the middle and lower plates (V23) is also \(10^{-2} \, \text{V}\).