Animated Solution for Physics - Electrostatics: A capacitor is made of two square plates each of side $a$ making a very small angle $\alpha$ between them, as shown in figure. The capacitance will be close to

Imagine you are looking at a standard parallel plate capacitor, but someone has slightly lifted one edge of the top plate. The plates are no longer perfectly parallel; instead, they form a very small angle $\alpha$ with each other. Because the separation distance is not constant, we cannot simply plug the values into our beloved formula $C=\frac{{\epsilon }_{0}A}{d}$.

To tackle this, we must rely on the power of calculus. We will slice the entire capacitor into infinitesimally thin strips, treating each strip as a tiny, perfect parallel plate capacitor.

Let's define our coordinate system. We place the origin at the left edge of the bottom plate. Now, consider a vertical strip of width $dx$ located at a distance $x$ from the left edge.

Since the plates are square with side length $a$, the area of this tiny strip is simply its length times its width:

Next, we need to find the separation between the plates at this exact position $x$. At the left edge ($x=0$), the separation is $d$. As we move to the right, the top plate rises. The additional height is given by basic trigonometry as $x\tan \alpha$. Because the angle $\alpha$ is extremely small, we can safely use the small-angle approximation $\tan \alpha \approx \alpha$.

Thus, the separation distance $y$ at position $x$ is:

The capacitance of our infinitesimally small strip, $dC$, is given by the standard formula applied to its tiny dimensions:

Since all these strips are connected across the same potential difference (from the left edge to the right edge), they are effectively in parallel. To find the total equivalent capacitance $C$, we simply integrate $dC$ from $x=0$ to $x=a$:

Evaluating this integral yields a natural logarithm:

We have a mathematically correct answer, but if you look at the options, there are no logarithms! This is where the physics intuition kicks in. The problem explicitly states that $\alpha$ is very small, which implies that the term $\frac{\alpha a}{d}$ is much less than 1.

Whenever we encounter $\ln (1+z)$ where $z\ll 1$, we must use the Taylor series expansion:

For our purposes, taking just the first two terms is sufficient to capture the first-order correction due to the tilt. Substituting $z=\frac{\alpha a}{d}$, we get:

Now, let's plug this approximation back into our capacitance expression:

To simplify, we can factor out $\frac{\alpha a}{d}$ from the terms inside the parenthesis:

The $\alpha$ in the numerator and denominator beautifully cancel out, leaving us with our final, elegant result:

This perfectly matches option (c). It's a brilliant demonstration of how calculus and algebraic approximations work hand-in-hand in physics!