Question

Two square plates with side length $l$ are arranged parallel to each other at distance d. They are charged to a potential difference $V_0$ and isolated from the source. A dielectric with a permittivity $\epsilon_r$ = 2 whose thickness is d and width is equal to that of the plates is drawn into the space between the plates. The length of the dielectric is greater than $l$ (figure). The resulting force F acting on the dielectric at distance $x = l/2$ is $\frac{\alpha CV_0^2}{\beta l}$ where $C = \frac{\epsilon_0 l^2}{d}$. Take $\alpha$ and $\beta$ as minimum integers. Find $\beta - \alpha$.

Answer 1

7

Answer 2

The capacitance of the parallel plate capacitor with a dielectric inserted to a distance $x$ is given by the sum of the capacitances of the part with dielectric and the part with air. The area of the part with dielectric is $l \times x$, and the area of the part with air is $l \times (l-x)$. The capacitance of the part with dielectric is $C_1 = \frac{\epsilon_r \epsilon_0 (lx)}{d}$. The capacitance of the part with air is $C_2 = \frac{\epsilon_0 l(l-x)}{d}$. The total capacitance is $C(x) = C_1 + C_2 = \frac{\epsilon_r \epsilon_0 lx}{d} + \frac{\epsilon_0 l(l-x)}{d} = \frac{\epsilon_0 l}{d} [\epsilon_r x + l - x]$. Given $\epsilon_r = 2$, so $C(x) = \frac{\epsilon_0 l}{d} [2x + l - x] = \frac{\epsilon_0 l}{d} (l+x)$. The capacitance of the empty capacitor is $C = \frac{\epsilon_0 l^2}{d}$. We can write $C(x) = \frac{\epsilon_0 l^2}{d} \frac{l+x}{l} = C \left( 1 + \frac{x}{l} \right)$.

The capacitor is charged to a potential difference $V_0$ and then isolated from the source. This means the charge $Q$ on the plates remains constant. The initial charge is $Q = C(0) V_0 = C V_0$. The energy stored in the capacitor is $U(x) = \frac{1}{2} \frac{Q^2}{C(x)}$. Substituting $Q = CV_0$ and $C(x) = C(1 + x/l)$, we get $U(x) = \frac{1}{2} \frac{(CV_0)^2}{C(1 + x/l)} = \frac{C^2 V_0^2}{2C(1 + x/l)} = \frac{C V_0^2}{2(1 + x/l)}$.

The force acting on the dielectric is given by $F = -\frac{dU}{dx}$ when the charge is constant. $F = -\frac{d}{dx} \left( \frac{C V_0^2}{2(1 + x/l)} \right) = -\frac{C V_0^2}{2} \frac{d}{dx} (1 + x/l)^{-1}$. Using the chain rule, $\frac{d}{dx} (1 + x/l)^{-1} = -1 (1 + x/l)^{-2} \times \frac{1}{l} = -\frac{1}{l(1 + x/l)^2}$. So, $F = -\frac{C V_0^2}{2} \left( -\frac{1}{l(1 + x/l)^2} \right) = \frac{C V_0^2}{2l(1 + x/l)^2}$.

We need to find the force at $x = l/2$. Substitute $x = l/2$ into the expression for $F(x)$: $F(l/2) = \frac{C V_0^2}{2l(1 + (l/2)/l)^2} = \frac{C V_0^2}{2l(1 + 1/2)^2} = \frac{C V_0^2}{2l(3/2)^2} = \frac{C V_0^2}{2l(9/4)} = \frac{C V_0^2}{9l/2} = \frac{2 C V_0^2}{9l}$.

The given form of the force is $\frac{\alpha CV_0^2}{\beta l}$. Comparing $\frac{2 CV_0^2}{9l}$ with $\frac{\alpha CV_0^2}{\beta l}$, we have $\frac{\alpha}{\beta} = \frac{2}{9}$. We are asked to find $\alpha$ and $\beta$ as minimum integers. Thus, $\alpha = 2$ and $\beta = 9$.

We need to find $\beta - \alpha$. $\beta - \alpha = 9 - 2 = 7$.