Capacitor Problem with distance-dependent dieletric

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SUMMARY

The discussion focuses on calculating the electric field, potential difference, capacitance, and work done when inserting a dielectric into a capacitor. The electric field within the dielectric is expressed as $$E = \frac{Q}{A \epsilon_o(1 + \frac{y}{d})}$$. The potential difference across the dielectric is given by $$\Delta V = \frac{Q}{A \epsilon_o} d\ln(2)$$, leading to a capacitance of $$C = \frac{A \epsilon_o}{d \ln2}$$. The work done to insert the dielectric is represented by the magnitude $$W=|U_{f}-U_{i}|=\left |\frac{Q^2}{2C_f}-\frac{Q^2}{2C_i}\right|$$, highlighting the dependency on the capacitances before and after insertion.

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Homework Statement
Two plates of separation d and cross-section area A make a parallel capacitor. Capacitor is charged fully to charge Q. The battery is then disconnected and space between plates is filled with inhomogeneous dieletric material with dieletric permittivity of $$\epsilon = \epsilon_o(1 + \frac{y}{d})$$ where origin of y is on the bottom plate.
Relevant Equations
C = Q/V
a) Find Electric Field at any point in the dieletric in the terms of the parameters given
Making a pillbox Gaussian surface with one end in the conductor where E = 0 and the other end in the dieletric we have that $$\oint D \cdot dS = \rho_s A \implies D = \rho_s \implies E = \frac{Q}{A \epsilon_o(1 + \frac{y}{d})}$$

b) Calculate potential difference across dieletric
$$\Delta V = - \int_{0}^{d} E \cdot dL = \int_{0}^{d} Edy = \int_{0}^{d} \frac{Q dy}{A \epsilon_o(1 + \frac{y}{d})} = \frac{Q}{A \epsilon_o} d\ln(2) $$

c) Calculate capacitance of the capacitor
$$C = \frac{Q}{\Delta V} = \frac{A \epsilon_o}{d \ln2}$$

d) How much work is done to insert the dieletric

I have no idea how to do this, my guess would be it would be equal to $$W = \frac{Q^2}{2C}$$ because that is the energy stored in a capacitor? Can anybody verify what I have done is correct?
 
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Parts (a)-(c) look fine. For part (d) you can calculate the magnitude but not the sign of the "work done" because the problem does not specify what entity is doing the work, the electrical forces or the external agent who inserts the dielectric. One is the negative of the other. In any case the magnitude of the work is $$W=|U_{f}-U_{i}|=\left |\frac{Q^2}{2C_f}-\frac{Q^2}{2C_i}\right|$$where ##C_f## and ##C_i## are, respectively, the capacitances after and before insertion of the dielectric.
 
Last edited:
kuruman said:
Parts (a)-(c) look fine. For part (d) you can calculate the magnitude but not the sign of the "work done" because the problem does not specify what entity is doing the work, the electrical forces or the external agent who inserts the dielectric. One is the negative of the other. In any case the magnitude is of the work is $$W=|U_{f}-U_{i}|=\left |\frac{Q^2}{2C_f}-\frac{Q^2}{2C_i}\right|$$where ##C_f## and ##C_i## are, respectively, the capacitances after and before insertion of the dielectric.
That makes complete sense thank you.
 

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