Problem 7 - College Physics

Question

(a) At one-third the original distance from a positive point charge, by what factor is the electric potential changed: \((1) 1 / 3,(2) 3,(3) 1 / 9,\) or (4) \(9 ?\) Why? (b) How far from \(a+1.0-\mu C\) charge is a point with an electric potential value of \(10 \mathrm{kV} ?\) (c) How much of a change in potential would occur if the point were moved to three times that distance?

Step-by-Step Solution

Answer

(a) The potential changes by a factor of 3. (b) The distance is 0.899 meters. (c) The potential change is 6666.67 volts.

Step 1: Understanding Electric Potential Change

(a) To determine how the electric potential changes when the distance from a point charge is altered, consider the formula for electric potential due to a point charge: \( V = k \frac{Q}{r} \), where \( V \) is the potential, \( k \) is Coulomb's constant, \( Q \) is the charge, and \( r \) is the distance from the charge. When the distance becomes one-third of the original, the new distance \( r' = \frac{r}{3} \). Substituting \( r' \) into the formula gives \( V' = k \frac{Q}{\frac{r}{3}} = 3k \frac{Q}{r} = 3V \). Thus, the factor by which the electric potential changes is 3. The correct answer is (2) 3.

Step 2: Calculating Distance for Given Potential Value

(b) To find the distance \( r \) from a \( +1.0\, \mu C \) charge where the electric potential is \( 10\, kV \), use the formula for electric potential: \( V = k \frac{Q}{r} \). Rearrange the formula to find \( r \): \( r = k \frac{Q}{V} \). Substitute \( Q = 1.0 \times 10^{-6} \) C, \( V = 10,000 \) V, and \( k = 8.99 \times 10^{9} \) N m²/C²: \( r = \frac{8.99 \times 10^{9} \times 1.0 \times 10^{-6}}{10,000} = 0.899 \) meters. Therefore, the point is 0.899 meters away from the charge.

Step 3: Finding Change in Potential at Triple Distance

(c) To calculate the change in potential when moving to three times the distance, use the initial results: Initially, \( V = k \frac{Q}{r} = 10,000 \) V and \( r = 0.899 \) m. If \( r' = 3 \times 0.899 \), then the new potential \( V' = k \frac{Q}{3r} = \frac{1}{3} \times 10,000 = 3,333.33 \) V. The change in potential is \( 10,000 - 3,333.33 = 6,666.67 \) V. The potential decreases by 6666.67 volts.

Key Concepts

point charge

A point charge is a charge that is concentrated at a single point in space. Imagine it like a tiny charged particle with no size or volume, just charge. In physics, when we say point charge, we usually consider problems where this assumption of charge concentrated in a point helps simplify calculations. This means everything about the electric potential and electric field can be derived assuming the charge acts just from that tiny point. This simplifies things because it allows us to use formulas like Coulomb's Law easily, considering only distance as important measure.

Coulomb's law

Coulomb's Law is a fundamental principle that helps us understand how electric forces work. It describes the force between two point charges. The formula is\[ F = k \frac{|Q_1 Q_2|}{r^2} \]where:

• \( F \) is the force between the charges,
• \( k \) is Coulomb's constant,
• \( Q_1 \) and \( Q_2 \) are the magnitudes of the charges,
• \( r \) is the distance between them.

Coulomb's law states that the force is proportional to the product of the two charges and inversely proportional to the square of the distance between them. This means as the distance increases, the force gets weaker, and as the charge values increase, the force gets stronger.

distance from charge

Distance from a charge is essential when calculating the electric potential and force exerted by that charge. In the context of point charges, the further you move from the charge, the weaker its electric field and potential effects become.Mathematically, in equations like those derived from Coulomb's Law and the electric potential formula, the distance from the charge is represented by \( r \). As a rule of thumb, when the distance \( r \) decreases, the effects (either force or potential) increase, and vice versa. For example, if you halve the distance to the charge, the force exerted by the charge will increase by four times (as per Coulomb's law where force is inversely related to the square of the distance). This relationship helps in predicting how electric potential changes with variation in distance.

electric potential formula

The electric potential formula is a key expression to determine the potential due to a point charge. This is given as:\[ V = k \frac{Q}{r} \]Here:

• \( V \) is the electric potential,
• \( k \) is Coulomb's constant,
• \( Q \) is the charge,
• \( r \) is the distance from the charge.

This formula shows that electric potential is directly proportional to the magnitude of the charge and inversely proportional to the distance \( r \) from the charge. This relationship implies that as you get closer to the charge, the potential increases, and as you move away, it decreases. Knowing how to use this formula helps determine how potential varies with distance, and if you know the potential, you can calculate distances to the charge, or vice versa.