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A \(5.00 \mathrm{nF}\) parallel-plate capacitor contains \(25.0 \mu \mathrm{J}\) of stored energy. (a) What is the potential difference across the capacitor? (b) By how many volts would you have to increase this potential difference in order for the capacitor to store \(50.0 \mu \mathrm{J}\) of potential energy?

Short Answer

Expert verified
(a) 100 V; (b) Increase by 41.42 V.

Step by step solution

01

Understand the Given Values and Formulas

We are given a parallel-plate capacitor with a capacitance \( C = 5.00 \mathrm{nF} = 5.00 \times 10^{-9} \mathrm{F} \) and stored energy \( U_1 = 25.0 \mu\mathrm{J} = 25.0 \times 10^{-6} \mathrm{J} \). We need to find the potential difference \( V_1 \) across the capacitor. We use the formula for energy stored in a capacitor: \( U = \frac{1}{2}CV^2 \).
02

Solve for Potential Difference \( V_1 \)

Rearrange the energy formula to solve for the potential difference: \( V_1 = \sqrt{\frac{2U_1}{C}} \). Substitute the given values: \( V_1 = \sqrt{\frac{2 \times 25.0 \times 10^{-6} \mathrm{J}}{5.00 \times 10^{-9} \mathrm{F}}} \). Calculate \( V_1 \):\[V_1 = \sqrt{\frac{50 \times 10^{-6}}{5 \times 10^{-9}}}= \sqrt{10^4}\= 100 \mathrm{V}\]{
03

Determine New Potential Energy and Voltage Difference

The new stored energy \( U_2 \) is \( 50.0 \mu\mathrm{J} = 50.0 \times 10^{-6} \mathrm{J} \). Use the formula \( U_2 = \frac{1}{2}C V_2^2 \) to find the new potential difference \( V_2 \). Plug in the values: \( V_2 = \sqrt{\frac{2U_2}{C}} \). Calculate \( V_2 \):\[V_2 = \sqrt{\frac{2 \times 50.0 \times 10^{-6}}{5.00 \times 10^{-9}}} \]\[= \sqrt{20 \times 10^{4}} = 141.42 \mathrm{V} \]
04

Calculate Voltage Increase Required

The voltage increase needed is \( \Delta V = V_2 - V_1 \). Substitute in the values found:\( \Delta V = 141.42 \mathrm{V} - 100 \mathrm{V} = 41.42 \mathrm{V} \).

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Key Concepts

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

Parallel-Plate Capacitor
A parallel-plate capacitor is a type of capacitor that consists of two conductive plates separated by a small distance. The space between the plates can either be a vacuum or filled with an insulating material known as a dielectric. Capacitors store electrical energy in the electric field created between these plates when a voltage is applied.
This simple yet fundamental electronic component is widely used in electronic circuits to store and release energy, filter signals, and perform other crucial tasks. The capacitance of a parallel-plate capacitor is determined by the area of the plates and the separation distance between them. The formula for capacitance \( C \) is given by:
  • \( C = \frac{\varepsilon A}{d} \)
where \( \varepsilon \) is the permittivity of the dielectric material, \( A \) is the area of one of the plates, and \( d \) is the separation distance between the plates. This formula highlights how larger plate areas and smaller separation distances lead to greater capacitance.
Potential Difference
Potential difference, often referred to as voltage, is the measure of electric potential energy per unit charge between two points in a circuit. In the context of a capacitor, it refers to the energy needed to move a charge between the two plates of the capacitor.
The potential difference across a capacitor can be calculated if the capacitance and the stored energy are known. The energy \( U \) stored in a capacitor is given by the formula:
  • \( U = \frac{1}{2}C V^2 \)
Rearranging this formula allows us to find the potential difference \( V \) when the energy and capacitance are known:
  • \( V = \sqrt{\frac{2U}{C}} \)
This equation helps us understand how the stored energy relates to the potential difference in a capacitor.
Electrical Potential Energy
Electrical potential energy is the energy stored in a system of charged particles due to their positions. In a parallel-plate capacitor, this energy is stored within the electric field between the plates.
For capacitors, this energy is described by the formula \( U = \frac{1}{2} C V^2 \), where \( U \) represents the potential energy, \( C \) the capacitance, and \( V \) the potential difference across the capacitor.
This equation tells us that the energy stored in a capacitor is directly proportional to its capacitance and the square of the voltage. Understanding electrical potential energy is crucial for controlling how capacitors are used in circuits, especially when it comes to storing and managing energy efficiently. It’s essential to ensure that the potential energy does not exceed the capacitor's limits, as this can result in a failure of the component.
Voltage Increase Calculation
Calculating a voltage increase in a parallel-plate capacitor involves determining the change in potential difference required to reach a specific level of stored energy. In the given exercise, we were asked to find out by how many volts we need to increase the potential difference to store more energy.
The first step is to find the new potential difference \( V_2 \) using the equation for potential energy \( U = \frac{1}{2} C V^2 \) by rearranging it to \( V = \sqrt{\frac{2U}{C}} \). Once \( V_2 \) is calculated, subtract the initial potential difference \( V_1 \) to find the voltage increase \( \Delta V = V_2 - V_1 \).
It's vital to carefully carry out these calculations to ensure the capacitor can safely handle the increased voltage, keeping in mind its voltage rating to prevent any damage.

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Most popular questions from this chapter

A uniform electric field has magnitude \(E\) and is directed in the negative \(x\) direction. The potential difference between point \(a\) (at \(x=0.60 \mathrm{~m}\) ) and point \(b\) (at \(x=0.90 \mathrm{~m}\) ) is \(240 \mathrm{~V}\). (a) Which point, \(a\) or \(b\), is at the higher potential? (b) Calculate the value of \(E\). (c) A negative point charge \(q=-0.200 \mu \mathrm{C}\) is moved from \(b\) to \(a\). Calculate the work done on the point charge by the electric field.

(a) How much charge does a battery have to supply to a \(5.0 \mu \mathrm{F}\) capacitor to create a potential difference of \(1.5 \mathrm{~V}\) across its plates? How much energy is stored in the capacitor in this case? (b) How much charge would the battery have to supply to store \(1.0 \mathrm{~J}\) of energy in the capacitor? What would be the potential across the capacitor in that case?

Two very large metal parallel plates that are \(25 \mathrm{~cm}\) apart, oriented perpendicular to a sheet of paper, are connected across the terminals of a \(50.0 \mathrm{~V}\) battery. (a) Draw to scale the lines where the equipotential surfaces due to these plates intersect the paper. limit your drawing to the region between the plates, avoiding their edges, and draw the lines for surfaces that are \(10.0 \mathrm{~V}\) apart, starting at the lowpotential plate. (b) These surfaces are separated equally in potential. Are they also separated equally in distance? (c) In words, describe the shape and orientation of the surfaces you just found.

For a particular experiment, helium ions are to be given a kinetic energy of \(3.0 \mathrm{MeV}\). What should the voltage at the center of the accelerator be, assuming that the ions start essentially at rest? A. \(-3.0 \mathrm{MV}\) B. \(+3.0 \mathrm{MV}\) C. \(+1.5 \mathrm{MV}\) D. \(+1.0 \mathrm{MV}\)

(a) If a spherical raindrop of radius \(0.650 \mathrm{~mm}\) carries a charge of \(-1.20 \mathrm{pC}\) uniformly distributed over its volume, what is the potential at its surface? (Take the potential to be zero at an infinite distance from the raindrop.) (b) Two identical raindrops, each with radius and charge specified in part (a), collide and merge into one larger raindrop. What is the radius of this larger drop, and what is the potential at its surface, if its charge is uniformly distributed over its volume?

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