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(I) A point charge \(Q\) creates an electric potential of \(+185 \mathrm{~V}\) at a distance of \(15 \mathrm{~cm}\). What is \(Q\) (let \(V=0\) at \(r=\infty\) )?

Short Answer

Expert verified
The charge \(Q\) is approximately \(3.08\, \text{nC}\).

Step by step solution

01

Understand the Problem

We need to calculate the charge \(Q\) that creates an electric potential of \(+185\, \mathrm{V}\) at a distance of \(15\, \mathrm{cm}\). The potential \(V\) at infinity is zero.
02

Write Down the Formula

The electric potential \(V\) due to a point charge \(Q\) at a distance \(r\) from the charge is given by the formula:\[ V = \frac{kQ}{r} \]where \(k = 8.988 \times 10^9 \, \mathrm{N\,m^2/C^2}\) is the Coulomb's constant.
03

Convert Units

Ensure all units are consistent. Convert the distance \(r\) from centimeters to meters: \[ r = 15 \, \text{cm} = 0.15 \, \text{m} \]
04

Substitute Values into the Formula

Using the given values in the formula:\[ 185 = \frac{(8.988 \times 10^9)Q}{0.15} \]Our goal is to solve for \(Q\).
05

Solve for Q

Rearrange the equation to solve for \(Q\):\[ Q = \frac{185 \times 0.15}{8.988 \times 10^9} \]Calculate \(Q\): \[ Q \approx 3.08 \times 10^{-9} \, \mathrm{C} \]
06

Conclude the Solution

The charge \(Q\) is approximately \(3.08 \, \text{nC}\) (nanoCoulombs). This is the charge responsible for the given electric potential at the specified distance.

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Key Concepts

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

Coulomb's Constant
Coulomb's constant, denoted as \( k \), is a key factor in calculating the electric potential and electric forces between point charges. It appears in Coulomb's Law and is crucial in the formula for electric potential, \( V = \frac{kQ}{r} \).
This constant has a value of \( 8.988 \times 10^9 \, \mathrm{N\,m^2/C^2} \), which shows how strongly two charges will interact with each other over a distance.
Think of \( k \) as a scaling factor that determines the strength of the electric force in a vacuum. Aiding in calculations involving electric fields, \( k \) ensures that we can deduce the impact of any charge at a representative distance, enhancing the predictability of electric interactions in physics.
Point Charge
A point charge is a simplified model of a charged particle that allows us to easily calculate and predict electric forces and fields. It's considered to have size zero, meaning all the charge is concentrated at one point.
This concept is extremely useful in theoretical calculations, as it provides a straightforward way to apply Coulomb's Law, which calculates force, and the electric potential formula. By assuming a charge is a point charge, we simplify complex systems into manageable mathematical equations.
  • Easy to model: Reduces complex charge distributions into simple single-point models.
  • Effective for theoretical work: Handy for predictions and understanding fundamental electric interactions.
In practical terms, several real-world charges can be approximated as point charges, especially when calculating forces or potentials over relatively large distances compared to the size of the charged body.
Electric Charge Calculation
Electric charge calculation involves determining the value of charge \( Q \) that induces a specific electric potential at a certain distance. In our exercise, this calculation uses the formula \( V = \frac{kQ}{r} \), where \( V \) is the potential, \( k \) is Coulomb's constant, and \( r \) is the distance from the charge.
To find \( Q \), you rearrange the formula to solve for it, taking care to ensure all units are consistent, such as converting centimeters to meters. The correct unit conversion ensures that distance is compatible with the values in the formula, leading to an accurate calculation.
  • Practical application: Electric charge calculations reveal how much charge a particle contains based on the potential it creates.
  • Systematic approach: Carefully solving step-by-step ensures clarity and accuracy in results.
Accurate calculations are crucial for understanding the behavior of charges and are frequently used in designing circuits and electrostatic applications.

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Most popular questions from this chapter

(II) A manufacturer claims that a carpet will not generate more than \(5.0 \mathrm{kV}\) of static electricity. What magnitude of charge would have to be transferred between a carpet and a shoe for there to be a \(5.0-\mathrm{kV}\) potential difference between the shoe and the carpet, approximating the shoe and the carpet as large sheets of charge separated by a distance \(d=1.0 \mathrm{~mm} ?\)

The potential in a region of space is given by \(V=B /\left(x^{2}+R^{2}\right)^{2}\) where \(B=150 \mathrm{~V} \cdot \mathrm{m}^{4}\) and \(R=0.20 \mathrm{~m}\) (a) Find \(V\) at \(x=0.20 \mathrm{~m}\) (b) Find \(\overrightarrow{\mathbf{E}}\) as a function of \(x\). (c) Find \(\overrightarrow{\mathbf{E}}\) at \(x=0.20 \mathrm{~m}\)

(I) Show that the electric field of a single point charge (the below equation) follows from Eq. \(5, V=\left(1 / 4 \pi \epsilon_{0}\right)(Q / r) .\) $$\begin{aligned} E &=k \frac{Q}{r^{2}} \\ E &=\frac{1}{4 \pi \epsilon_{0}} \frac{Q}{r^{2}} \end{aligned}$$

A lightning flash transfers \(4.0 \mathrm{C}\) of charge and \(4.8 \mathrm{MJ}\) of energy to the Earth. ( \(a\) ) Between what potential difference did it travel? (b) How much water could this energy boil, starting from room temperature?

(III) Electrons are accelerated by \(6.0 \mathrm{kV}\) in a CRT. The screen is \(28 \mathrm{~cm}\) wide and is \(34 \mathrm{~cm}\) from the \(2.6-\mathrm{cm}\) -long deflection plates. Over what range must the horizontally deflecting electric field vary to sweep the beam fully across the screen?

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