The charge on a capacitor plate in a circuit as a function of time is shown in the figure. What is t

Q: Why do we find the slope instead of just dividing charge by time?

Dividing total charge by total time ($q/t$) gives the *average* current over that interval. To find the *instantaneous* current at a specific moment (like $t=4\text{ s}$), we must use the derivative $\frac{dq}{dt}$, which corresponds to the slope of the graph at that exact point.

Q: What does a horizontal line on a $q-t$ graph physically mean?

A horizontal line means the charge is constant over time. If the charge isn't changing, no new charge is flowing onto or off the capacitor plate, which means the current is zero.

Q: How would I find the current at $t=1\text{ s}$?

At $t=1\text{ s}$, the graph is a straight line starting from $(0,0)$ to $(2,3)$. The slope is $\frac{3 - 0}{2 - 0} = 1.5$. So, the current would be $1.5\mu\text{A}$.

Q: Is the current constant throughout the entire $8\text{ s}$?

No! The slope changes. From $0$ to $2\text{ s}$, the current is $1.5\mu\text{A}$. From $2$ to $6\text{ s}$, it is $0$. From $6$ to $8\text{ s}$, the slope is $\frac{6 - 3}{8 - 6} = 1.5\mu\text{A}$ again.

Q: Do I need to convert microcoulombs to coulombs?

Not necessarily. Since the Y-axis is in microcoulombs ($\mu\text{C}$) and the X-axis is in seconds ($\text{s}$), the slope will naturally have units of $\mu\text{C/s}$, which is exactly microamperes ($\mu\text{A}$). This matches the given options perfectly.

Question

The charge on a capacitor plate in a circuit as a function of time is shown in the figure.

What is the value of current at $t = 4	ext{ s}$?

Options:

(A) $2\mu	ext{A}$

(B) $1.5\mu	ext{A}$

(C) Zero

(D) $3\mu	ext{A}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup
Imagine you are monitoring a capacitor in an electrical circuit. The graph provided gives us a direct visual representation of how the charge $q$ on the capacitor plate changes as time $t$ ticks forward. Our mission is to determine the electric current flowing through the circuit at the exact moment when $t = 4	ext{ s}$.

### The Master Equation
To solve this, we must bridge the gap between charge and current. What exactly is electric current? By definition, current $I$ is the rate at which electric charge flows. Mathematically, this is expressed as the time derivative of charge:
$$I = \frac{dq}{dt}$$

This elegant equation is our master key. It tells us that if we know how charge changes with time, we can instantly find the current.

### Graphical Interpretation
Now, let's translate this calculus concept into geometry. In a graph where charge $q$ is plotted on the Y-axis and time $t$ is on the X-axis, the derivative $\frac{dq}{dt}$ represents the **slope** of the graph at any given point.

Therefore, finding the current at $t = 4	ext{ s}$ is equivalent to finding the slope of the $q-t$ graph at $t = 4	ext{ s}$.

### Final Calculation
Let's direct our attention to the graph at $t = 4	ext{ s}$. Notice the segment of the graph between $t = 2	ext{ s}$ and $t = 6	ext{ s}$. In this entire interval, the graph is a perfectly horizontal line, meaning the charge is constant at $3\mu	ext{C}$.

What is the slope of a horizontal line? It is exactly zero! Since the charge is not changing, the rate of change of charge is zero.
$$\frac{dq}{dt} = 0$$

Consequently, the current at $t = 4	ext{ s}$ is **zero**. The beauty of graphical analysis is that it often saves us from complex calculations, giving us the answer through simple visual inspection.

Animated Solution for Physics - Electrostatics: The charge on a capacitor plate in a circuit as a function of time is shown in the figure. What is the value of current at $t = 4 s$ ?

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Visualized Solution

$Analyzing the Graph$

$Current and Charge$

$Graphical Meaning$

$Slope at t = 4 s$

$Final Answer$

$The Way Forward$

The Sigma Insight: Capacitance and Capacitors

Analyzing the Setup Imagine you are monitoring a capacitor in an electrical circuit

The Master Equation To solve this, we must bridge the gap between charge and current

Graphical Interpretation Now, let's translate this calculus concept into geometry

Final Calculation Let's direct our attention to the graph at $t = 4 s$

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Visualized Solution

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

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

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Slope at t=4 s

.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