archaic
- 688
- 214
- Homework Statement
- We have a capacitor made of two square plates of length ##\ell##, and separated by a distance ##d##. Charges ##+Q## and ##-Q## are placed on the plates.
A material of dielectric constant ##\kappa=4.50## is inserted a distance ##x## into the capacitor as shown in the figure below.
The system can be considered as two capacitors connected in parallel.
a) Find the equivalent capacitance.
b) Calculate the energy stored in the capacitor.
c) Find the direction and magnitude of the force exerted by the plates on the dielectric.
b) If the capacitor was charge to ##2.20\times10^3## V before the dielectric was inserted, then give a numerical value of the force in micro newtons if ##x=\ell/2##, ##\ell=5.10## cm, and ##d=1.80## mm.
- Relevant Equations
- $$C=\frac{Q}{\Delta V}$$
a)$$C=\frac{\kappa\epsilon_0\ell x}{d}+\frac{\epsilon_0\ell(\ell-x)}{d}=\frac{\epsilon _0ℓ\left(\kappa x+ℓ-x\right)}{d}$$
b)$$U=\frac{1}{2C}Q^2=\frac{dQ^2}{2\epsilon _0ℓ\left(\kappa x+l-x\right)}$$
c)$$F=-\frac{dU}{dx}=\frac{dQ^2\left(\kappa -1\right)}{2\epsilon _0ℓ\left(\kappa x+ℓ-x\right)^2}$$and it is directed to the right, since it is positive.
b) I first find the charge using ##Q=C_0\Delta V=\frac{\epsilon_0\ell^2\Delta V}{d}##$$\begin{align*}
F&=\frac{d\left(\kappa -1\right)}{2\epsilon _0ℓ\left(\kappa\fracℓ2+ℓ-\fracℓ2\right)^2}\frac{\epsilon_0^2\ell^4(\Delta V)^2}{d^2}\\
&=\frac{d\left(\kappa -1\right)}{2\epsilon _0\frac{ℓ^3}{4}\left(\kappa+1\right)^2}\frac{\epsilon_0^2\ell^4(\Delta V)^2}{d^2}\\
&=\frac{2\epsilon_0\ell(\Delta V)^2(\kappa-1)}{d(\kappa+1)^2}\\
&=\frac{2(5.10\times10^{-2})(2.20\times10^3)^2(4.50-1.00)}{(4\pi\times8.99\times10^9)(1.80\times10^{-3})(4.50+1)^2}\\
&=0.000281\,N=281\,\mu N
\end{align*}$$