Two square metal plates of side 1 m are kept 0.01 m apart like a parallel plate capacitor in air in

Q: Why are the two parts considered to be in parallel?

Because the top edges of both the air part and the oil part are at the same potential (connected to the positive terminal), and the bottom edges are at the same potential (connected to the negative terminal).

Q: Why does the current flow even though the battery voltage is constant?

As the plates are lowered, the dielectric replaces air, increasing the total capacitance. Since $q = CV$, an increasing capacitance at a constant voltage requires more charge to flow from the battery, which constitutes a current.

Q: What is $\frac{dx}{dt}$ in this context?

It is the rate at which the submerged depth $x$ is increasing, which is exactly the speed at which the plates are being lowered into the oil.

Q: Does the distance between the plates $d$ change?

No, the plates are rigid and kept at a constant separation of $0.01\text{ m}$. Only the dielectric medium between them is changing.

Q: What if the plates were disconnected from the battery before lowering?

If disconnected, the charge $q$ would remain constant. As capacitance $C$ increases, the potential difference $V = q/C$ across the plates would decrease, and no current would flow from the battery.

Question

Two square metal plates of side $1	ext{ m}$ are kept $0.01	ext{ m}$ apart like a parallel plate capacitor in air in such a way that one of their edges is perpendicular to an oil surface in a tank filled with an insulating oil. The plates are connected to a battery of emf $500	ext{ V}$. The plates are then lowered vertically into the oil at a speed of $0.001	ext{ ms}^{-1}$. Calculate the current drawn from the battery during the process. (Dielectric constant of oil $= 11$, $\varepsilon_0 = 8.85 	imes 10^{-12}	ext{ C}^2	ext{N}^{-1}	ext{m}^{-2}$).

Sigma Percentile · Accepted Answer

### Analyzing the Setup
Imagine a square parallel plate capacitor being slowly lowered into a tank of oil. As it dips, the oil replaces the air between the plates. We can think of this setup as two separate capacitors connected in parallel: one part still in the air, and the other part submerged in the oil.

Since they are connected to the same battery, they are in parallel. The total capacitance is simply the sum of the two. And remember, the current drawn from the battery is just the rate at which the total charge on these plates changes with time.

### The Master Equation
Let's set up our variables. The plates are squares of side $a$, separated by a distance $d$. If they are submerged to a depth $x$, the area of the plates in the oil is $ax$, and the remaining area in the air is $a(a - x)$.

Now, let's write down the capacitance for each part. For the part in the air, $C_1$ is $\varepsilon_0$ times its area, $a(a - x)$, divided by $d$. For the submerged part, $C_2$, we must include the dielectric constant $K$ of the oil.

$$C_1 = \frac{\varepsilon_0 a(a - x)}{d}$$
$$C_2 = \frac{K \varepsilon_0 a x}{d}$$

Adding them up gives us the total equivalent capacitance. Notice how we can factor out the common terms to simplify the expression. The total capacitance depends linearly on the submerged depth $x$.

$$C = \frac{\varepsilon_0 a(a - x)}{d} + \frac{K \varepsilon_0 a x}{d} = \frac{\varepsilon_0 a}{d} [a + (K - 1)x]$$

The total charge $q$ stored on the plates is the total capacitance $C$ multiplied by the constant battery voltage $V$.

$$q = CV = \frac{\varepsilon_0 a V}{d} [a + (K - 1)x]$$

To find the current $i$, we differentiate the charge $q$ with respect to time $t$. Since the voltage, dimensions, and dielectric constant are all constants, the only thing changing with time is the depth $x$. The derivative $\frac{dx}{dt}$ is simply the speed at which the plates are being lowered.

$$i = \frac{dq}{dt} = \frac{\varepsilon_0 a V}{d} (K - 1) \frac{dx}{dt}$$

### Final Calculation
We have our master equation! Now, let's carefully substitute all the given values. The side $a$ is $1	ext{ m}$, distance $d$ is $0.01	ext{ m}$, voltage is $500	ext{ V}$, $K$ is $11$, and the speed $\frac{dx}{dt}$ is $0.001	ext{ ms}^{-1}$.

$$i = \frac{8.85 	imes 10^{-12} 	imes 1 	imes 500}{0.01} (11 - 1) (0.001)$$

Calculating this gives us the final current.

$$i = 44.25 	imes 10^{-10}	ext{ A} = 4.43 	imes 10^{-9}	ext{ A}$$

**The current drawn from the battery is $4.43 	imes 10^{-9}	ext{ A}$.** A very tiny current, but physically significant!

Visualized Solution

$Visualizing the System$

$Equivalent Capacitance$

$Capacitance Expressions$

$Total Capacitance$

$Charge on the Capacitor$

$Current Calculation$

$Substituting Values$

$Final Answer$

$The Way Forward$

The Sigma Insight: Capacitance and Capacitors

Analyzing the Setup Imagine a square parallel plate capacitor being slowly lowered into a tank of oil

The Master Equation Let's set up our variables

Final Calculation We have our master equation! Now, let's carefully substitute all the given values

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Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Visualizing the System

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Equivalent Capacitance

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Capacitance Expressions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Total Capacitance

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Charge on the Capacitor

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Current Calculation

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Substituting Values

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Final Answer

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The Way Forward