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Two 10 -cm-diameter electrodes \(0.50 \mathrm{cm}\) apart form a parallel-plate capacitor. The electrodes are attached by metal wires to the terminals of a \(15 \mathrm{V}\) battery. What are the charge on each electrode, the electric field strength inside the capacitor, and the potential difference between the electrodes a. While the capacitor is attached to the battery? b. After insulating handles are used to pull the electrodes away from each other until they are \(1.0 \mathrm{cm}\) apart? The electrodes remain connected to the battery during this process. c. After the original electrodes (not the modified electrodes of part b) are expanded until they are \(20 \mathrm{cm}\) in diameter while remaining connected to the battery?

Short Answer

Expert verified
Recalculate the values of charge, electric field and potential difference for each change in the configuration as described above using appropriate formulas. For final results, substitute the numerical values of \(\varepsilon_0\), \(A\), \(d\), and \(V\) into these formulas and simplify.

Step by step solution

01

Compute the Capacitance

First, compute the initial capacitance using the formula for a parallel-plate capacitor \(C = \varepsilon_0(A/d)\). Here, the diameter of each electrode is 10 cm, which makes the radius \(r = 5 cm = 0.05 m\). So, the area \(A = \pi r^2 = \pi \times (0.05 m)^2\). The initial distance between the electrodes is 0.5 cm, which is \(d = 0.005 m\). Using these values, \(\varepsilon_0 = 8.85 \times 10^{-12} \, F/m\), you can find the capacitance \(C\).
02

Calculate the Initial Charge and Electric Field

With the capacitance and the battery voltage (15V), calculate the charge on the capacitor using \(Q = CV\). Also calculate the initial electric field using \(E = V/d\).
03

Find the Charge, Electric Field, Potential Difference after Increasing the Distance

When the distance between the electrodes is increased to 1 cm (or \(d = 0.01 m\)), the charges on the plates remain constant (as the battery stays connected), so \(Q\) is same as before. However, the electric field \(E\) will change and can be recalculated using \(E = V/d\). The potential difference also remains constant at 15V.
04

Compute the Capacitance, Charge, Electric Field, Potential Difference after Increasing the Diameter

When the diameter of the plates is increased to 20 cm (or radius \(r = 10 cm\)), both area \(A\) and capacitance \(C\) increase. Recalculate \(A = \pi r^2\) and \(C = \varepsilon_0(A/d)\). The charge \(Q\) changes (since \(Q = CV\)), so compute the new charge. As the potential difference (V) remains constant, the electric field reduces and is given by \(E = V/d\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Capacitance
Capacitance describes the ability of a capacitor to store charge. It is a key property for parallel-plate capacitors and is influenced by various factors.
In general, the capacitance of a parallel-plate capacitor is given by the formula:
  • \[ C = \varepsilon_0 \frac{A}{d} \]
where \( C \) is the capacitance, \( \varepsilon_0 \) is the permittivity of free space \((8.85 \times 10^{-12} \, F/m)\), \( A \) is the area of one of the plates, and \( d \) is the separation between the electrodes.
The larger the area \( A \), the more charge the plates can hold, increasing the capacitance. On the flip side, increasing the distance \( d \) between plates decreases the capacitance as the electric field has more space to spread out. By understanding this, we can appreciate how changes in the physical parameters of the capacitor affect its ability to store charge.
Electric Field
The electric field inside a capacitor represents the force that acts on charges within the device. It's a vital factor that determines the behavior of charges and the overall functioning of the capacitor.
This field is calculated using the formula:
  • \[ E = \frac{V}{d} \]
where \( E \) is the electric field strength, \( V \) is the potential difference (voltage), and \( d \) is the distance between the electrodes.
Hence, the electric field is directly proportional to the voltage and inversely proportional to the distance between the plates. If the voltage increases or the plate separation decreases, the electric field becomes stronger, meaning a greater force per unit charge. This concept is crucial for understanding how energy is distributed and why the field changes when plate separation or voltage is altered.
Potential Difference
The potential difference, or voltage, across a capacitor is a measure of energy per unit charge. It is an essential concept as it dictates the energy stored in a capacitor.
The potential difference is usually set by the external circuit, such as the battery connected to the capacitor. In our scenario, the potential difference remains constant at 15V because the battery maintains steady voltage across the capacitor.
This constant voltage ensures that when the separation between the plates changes, the potential difference doesn't change, though this results in an adjustment of the capacitance and electric field strength. Understanding how the potential difference remains constant while other properties change provides deeper insight into the flexible nature of capacitors in different configurations.
Charge Calculation
Calculating the charge on a capacitor involves using its capacitance and the voltage across its plates.
The relevant formula is:
  • \[ Q = C \cdot V \]
where \( Q \) is the charge, \( C \) is the capacitance, and \( V \) is the potential difference.
Understanding this relationship helps to determine how changes in capacitance influence the charge stored. For instance, when electrode size is altered, changing the capacitance, we can use this formula to find the resulting charge. While the battery is connected, the charge also depends on whether the configuration affects the overall potential difference. Thus, knowing these conditions is vital for accurate charge assessment.
Electrode Separation
Electrode separation is the gap between the two plates of a capacitor and is a critical parameter because it affects most properties of the capacitor.
Separation is denoted as \( d \) and features in key equations for capacitors, such as those for capacitance and electric field. By increasing the separation between the electrodes, the capacitance decreases as the electric field diminishes spreading across a larger space. Meanwhile, the potential difference, if controlled by a battery, remains the same, showcasing how separation influences only specific properties.
When insights on separation changes are paired with calculations of capacitance and electric fields, learners can grasp why objects like dielectric materials which decrease effective separation, dramatically alter capacitor behavior. Understanding this helps in predicting and controlling capacitor performance in circuits.

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Most popular questions from this chapter

A \(20 \mathrm{cm} \times 2.0 \mathrm{cm}\) parallel-plate capacitor has a \(2.0 \mathrm{mm}\) spacing. The electric field strength inside the capacitor is \(10 \times 10^{5} \mathrm{V} / \mathrm{m}\) a. What is the potential difference across the capacitor? b. How much charge is on each plate?

The electron gun in a TV picture tube accelerates electrons between two parallel plates \(1.2 \mathrm{cm}\) apart with a \(25 \mathrm{kV}\) potential difference between them. The electrons enter through a small bole in the negative plate, accelerate, then exit through a small hole in the positive plate. Assume that the holes are small enough not to affect the electric field or potential. a. What is the electric field strength between the plates? b. With what speed does an electron exit the electron gun if its entry speed is close to zero? Note \(\Rightarrow\) The exit speed is so fast that we really need to use the theory of relativity to compute an accurate value. Your answer to part b is in the right range but a little too big. 4

What is the speed of an electron that has been accelerated from rest through a potential difference of \(1000 \mathrm{V} ?\)

A proton is released from rest at the positive plate of a parallel-plate capacitor. It crosses the capacitor and reaches the negative plate with a speed of \(50,000 \mathrm{m} / \mathrm{s}\). What will be the proton's final speed if the experiment is repeated with double the amount of charge on each capacitor plate?

\(A-2.0\) nC charge and \(a+2.0\) nC charge are located on the \(x\) axis at \(x=-1.0 \mathrm{cm}\) and \(x=+1.0 \mathrm{cm},\) respectively. a. At what position or positions on the \(x\) -axis is the electric field zero? b. At what position or positions on the \(x\) -axis is the electric potential zero? c. Draw graphs of the electric field strength and the electric potential along the \(x\) -axis.

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