Let's solve the question step-by-step: --- ### Q13. A parallel-plate capacitor... #### (a) **Sketch a graph of voltage \( V \) as a function of \( x \) (with air between the plates)** **Given:** - Plate separation = \( d \) - Voltage across plates = \( V_{\text{bat}} \) - \( x = 0 \): positively charged plate - \( x = d \): negatively charged plate ##### **Step 1: Relationship between Voltage and Position** For a parallel-plate capacitor **with air** (i.e., vacuum or air as dielectric, uniform field): - Electric field, \( E = \frac{V_{\text{bat}}}{d} \) (constant) - Potential difference across a distance \( x \): \( V(x) = E \cdot x = \frac{V_{\text{bat}}}{d} x \) ##### **Step 2: Graph and Points** - At \( x = 0 \): \( V(0) = 0 \) - At \( x = d/4 \): \( V(d/4) = \frac{V_{\text{bat}}}{d} \cdot \frac{d}{4} = \frac{V_{\text{bat}}}{4} \) - At \( x = d/2 \): \( V(d/2) = \frac{V_{\text{bat}}}{d} \cdot \frac{d}{2} = \frac{V_{\text{bat}}}{2} \) - At \( x = 3d/4 \): \( V(3d/4) = \frac{3V_{\text{bat}}}{4} \) - At \( x = d \): \( V(d) = V_{\text{bat}} \) ##### **Step 3: Sketch** - **Graph**: Straight line from (0, 0) to (d, \( V_{\text{bat}} \)) - **Slope**: \( \frac{V_{\text{bat}}}{d} \) **Sketch (ASCII version):** ``` V | * | * | * |* +------------------- x 0 d/4 d/2 3d/4 d ``` Each * corresponds to the points above. --- #### (b) **Dielectric slab of thickness \( d/2 \) inserted** - The slab is centered at \( x = d/2 \) - Thickness = \( d/2 \) - So slab spans from \( x = d/4 \) to \( x = 3d/4 \) - The rest (air) are from \( 0 \) to \( d/4 \) and \( 3d/4 \) to \( d \) ##### **Step 1: Voltage distribution** For regions: - **Air**: \( E_1 = \frac{V_1}{d_1} \) - **Dielectric**: \( E_2 = \frac{V_2}{d_2} = \frac{V_2}{d/2} \) where \( V_2 \) is voltage drop across dielectric - **Relation**: \( E_2 = \frac{E_1}{\kappa} \), \( \kappa \) = dielectric constant ##### **Step 2: Total Voltage** Total voltage: \[ V_{\text{bat}} = V_{\text{air1}} + V_{\text{dielectric}} + V_{\text{air2}} \] Let \( V_1 \) = voltage across each air region and \( V_2 \) = voltage across dielectric. - Air regions: \( d_1 = d_3 = d/4 \) - Dielectric region: \( d_2 = d/2 \) Using \( E = V/d \): \[ E_1 = \frac{V_1}{d/4},\quad E_2 = \frac{V_2}{d/2} \] But \( E_2 = \frac{E_1}{\kappa} \). So, \[ V_{\text{bat}} = V_1 + V_2 + V_1 \] \[ \frac{V_1}{d/4} = E_1 \implies V_1 = E_1 \frac{d}{4} \] \[ V_2 = E_2 \frac{d}{2} = \frac{E_1}{\kappa} \frac{d}{2} \] So, \[ V_{\text{bat}} = 2V_1 + V_2 = 2E_1\frac{d}{4} + E_2\frac{d}{2} = E_1\frac{d}{2} + \frac{E_1}{\kappa} \frac{d}{2} \] \[ V_{\text{bat}} = \frac{d}{2} E_1 \left(1 + \frac{1}{\kappa}\right) \] \[ E_1 = \frac{2V_{\text{bat}}}{d \left(1 + \frac{1}{\kappa}\right)} \] Now, voltages: - For \( 0 < x < d/4 \): \( V(x) = E_1 x \) - For \( d/4 < x < 3d/4 \): \( V(x) = E_1 \frac{d}{4} + E_2(x - d/4) \) - For \( 3d/4 < x < d \): \( V(x) = E_1 \frac{d}{4} + E_2 \frac{d}{2} + E_1 (x - 3d/4) \) ##### **Step 3: Sketch** - **Slope in air**: Steeper (higher field) - **Slope in dielectric**: Shallower (lower field) **Graph:** - From \( x = 0 \) to \( d/4 \): steep linear rise - From \( x = d/4 \) to \( 3d/4 \): less steep linear rise - From \( x = 3d/4 \) to \( x = d \): steep linear rise **ASCII Sketch:** ``` V | | ____ | / | / |_/ +-------------------- x 0 d/4 3d/4 d ``` The voltage increases more slowly between \( d/4 \) and \( 3d/4 \) (inside the dielectric). --- ## **Summary Table:** | Point | (a) Air only | (b) With Dielectric Slab | |-----------|--------------------|-------------------------| | \( x=0 \) | 0 | 0 | | \( x=d/4 \) | \( V_{\text{bat}}/4 \) | \( V_1 \) | | \( x=d/2 \) | \( V_{\text{bat}}/2 \) | \( V_1 + V_2/2 \) | | \( x=3d/4 \) | \( 3V_{\text{bat}}/4 \) | \( V_1 + V_2 \) | | \( x=d \) | \( V_{\text{bat}} \) | \( V_{\text{bat}} \) | --- **Let me know if you want the explicit values for \( V_1 \) and \( V_2 \) in terms of \( V_{\text{bat}} \) and \( \kappa \)!**

**Final Answers:** **(a)** - The voltage \( V \) varies linearly from 0 at \( x=0 \) to \( V_{\text{bat}} \) at \( x=d \). - At \( x=0 \): \( V=0 \) - At \( x=d/4 \): \( V=V_{\text{bat}}/4 \) - At \( x=d/2 \): \( V=V_{\text{bat}}/2 \) - At \( x=3d/4 \): \( V=3V_{\text{bat}}/4 \) - At \( x=d \): \( V=V_{\text{bat}} \) *(Graph is a straight line from (0,0) to (d, \(V_{\text{bat}}\)))* --- **(b)** - With a dielectric slab of thickness \( d/2 \) centered at \( x=d/2 \): - The voltage distribution has two regions: - From \( x=0 \) to \( x=d/4 \): steep slope (air) - From \( x=d/4 \) to \( x=3d/4 \): gentler slope (dielectric) - From \( x=3d/4 \) to \( x=d \): steep slope again - The voltage at key points: - \( x=0 \): 0 - \( x=d/4 \): \( V_1 = \frac{2V_{\text{bat}}}{1 + 1/\kappa} \times \frac{1}{2} \) (since \( V_1 = \frac{V_{\text{bat}}}{2} \times \frac{1}{1 + 1/\kappa} \)) - \( x=d/2 \): \( V_{d/2} = V_1 + \frac{V_2}{2} \) - \( x=3d/4 \): \( V_1 + V_2 \) - \( x=d \): \( V_{\text{bat}} \) *(Graph shows two slopes: steeper in air regions, gentler in dielectric region)*