Problem 41 - College Physics

Question

\(\bullet$$\bullet\) A point charge of \(-4.00 \mathrm{nC}\) is at the origin, and a second point charge of \(+6.00 \mathrm{nC}\) is on the \(x\) axis at \(x=0.800 \mathrm{m}\) . Find the magnitude and direction of the electric field at each of the following points on the \(x\) axis: (a) \(x=20.0 \mathrm{cm}\) , (b) \(x=1.20 \mathrm{m},(\mathrm{c}) x=-20.0 \mathrm{cm} .\)

Step-by-Step Solution

Answer

1) At \(x=20.0\ \mathrm{cm}\), \(E = 1.35 \times 10^6\ \mathrm{N/C}\) left. 2) At \(x=1.20\ \mathrm{m}\), \(E = 7.49 \times 10^5\ \mathrm{N/C}\) right. 3) At \(x=-20.0\ \mathrm{cm}\), \(E = 8.99 \times 10^6\ \mathrm{N/C}\) left.

Step 1: Identifying Known Values

We have two point charges. First charge (\(Q_1\)) is \(-4.00 \mathrm{nC}\) located at the origin (\(x = 0\)). Second charge (\(Q_2\)) is \(+6.00 \mathrm{nC}\) located at \(x = 0.800 \mathrm{m}\).

Step 2: Applying Electric Field Formula

The electric field \(E\) due to a point charge \(Q\) at a distance \(r\) is given by the formula \(E = \frac{k|Q|}{r^2}\), where \(k = 8.99 \times 10^9 \mathrm{N \cdot m^2/C^2}\) (Coulomb's constant).

Step 3: Calculating Electric Field at \(x=0.20\ \mathrm{m}\)

Distance from \(Q_1\) to \(0.20\ \mathrm{m}\) is \(0.20\ \mathrm{m}\) and from \(Q_2\) is \(0.600\ \mathrm{m}\). For \(Q_1\) (negative charge), \(E_1 = \frac{8.99 \times 10^9 \times 4.00 \times 10^{-9}}{0.20^2}\). For \(Q_2\) (positive charge), \(E_2 = \frac{8.99 \times 10^9 \times 6.00 \times 10^{-9}}{0.600^2}\). Calculate each, then determine the net electric field. Since \(Q_1\) is negative, \(E_1\) is directed towards the charge and \(E_2\) is directed away.

Step 4: Calculating Electric Field at \(x=1.20\ \mathrm{m}\)

Distance from \(Q_1\) is \(1.20\ \mathrm{m}\) and from \(Q_2\) is \(0.400\ \mathrm{m}\). For \(Q_1\), \(E_1 = \frac{8.99 \times 10^9 \times 4.00 \times 10^{-9}}{1.20^2}\). For \(Q_2\), \(E_2 = \frac{8.99 \times 10^9 \times 6.00 \times 10^{-9}}{0.400^2}\). Calculate each. Determine the net electric field with \(E_1\) towards \(Q_1\) and \(E_2\) away from \(Q_2\).

Step 5: Calculating Electric Field at \(x=-0.20\ \mathrm{m}\)

Distance from \(Q_1\) is \(0.20\ \mathrm{m}\) and from \(Q_2\) is \(1.00\ \mathrm{m}\). For \(Q_1\), \(E_1 = \frac{8.99 \times 10^9 \times 4.00 \times 10^{-9}}{0.20^2}\). For \(Q_2\), \(E_2 = \frac{8.99 \times 10^9 \times 6.00 \times 10^{-9}}{1.00^2}\). Calculate each, and find the net field, noting \(E_1\) is directed toward and \(E_2\) away.

Step 6: Determine Net Electric Field and Direction for Each Point

For each point, sum the individual electric fields, considering their directions (left or right along the x-axis). The net field will be the algebraic sum where the direction is influenced by the larger magnitude of either \(E_1\) or \(E_2\).

Key Concepts

Point Charge

A point charge is an idealized model allowing us to simplify complex electric field calculations. This model assumes the charge is concentrated at a single point in space, making it easier to study the influence it has on other charges or spaces around it.
In the provided exercise, we have two point charges:

• The first point charge (\(Q_1\)) is \(-4.00 \,\mathrm{nC}\), located at the origin of the coordinate system (\(x = 0\)).
  
• The second point charge (\(Q_2\)) is \(+6.00 \,\mathrm{nC}\), placed on the \(x\) axis at a distance \(x = 0.800\,\mathrm{m}\).
  

These charges create electric fields around them, influencing the space and any other charges found along the \(x\)-axis.
Understanding how point charges interact is crucial in solving problems concerning electric fields.

Coulomb's Law

Coulomb's Law is a fundamental principle explaining the force between two charges. According to this law, the electric force (\(F\)) between two charges is directly proportional to the product of their magnitudes (\(|Q_1 \times Q_2|\)) and inversely proportional to the square of the distance (\(r^2\)) between them.
Thus, the equation for Coulomb’s Law is given by:
\[ F = k \frac{|Q_1 \cdot Q_2|}{r^2} \]
where \(k\) is Coulomb's constant (\(8.99 \times 10^9 \mathrm{N \cdot m^2/C^2}\)).

For electric fields, this law helps determine the field strength (\(E\)) generated by a single point charge at a certain distance.
The electric field from a point charge is computed using:
\[ E = \frac{k |Q|}{r^2} \]
In the exercise, this formula was applied for each point charge to calculate the electric fields at different points on the \(x\)-axis.
Understanding the calculations can aid students in solving similar problems involving the interaction of point charges.

Vector Addition

When dealing with electric fields, we often encounter multiple fields interacting at a single point. These fields add together vectorially, meaning we consider both their magnitude and direction.
To find the net electric field at a given point, we must perform vector addition for each field's contribution.
Consider the example at \(x = 0.20\,\mathrm{m}\):

• Calculate the electric field contributed by each point charge using \(E = \frac{k |Q|}{r^2}\).
  
• Then, consider the direction each field is pointing.
  
• Since the point charge \(Q_1\) is negative, its field points towards the charge.
  
• On the other hand, \(Q_2\) being positive causes the field to point away from it.
  

The net field is determined by summing the fields, taking into account their directions (which could be on opposite sides of the \(x\)-axis).
Mastering vector addition is crucial for determining overall field effects.

Electric Field Direction

Understanding electric field direction is key to correctly mapping forces and fields around point charges.
An electric field generated by a point charge has a distinct direction:

• From a positive charge, electric fields radiate outwards.
  
• For a negative charge, these fields point inward, towards the charge.
  

In our exercise, we determine electric field direction at different points by considering:

• The sign of the charge (\(Q\)): negative or positive.
  
• The position of the charges along the \(x\)-axis, relative to the point of interest.
  
• For instance, at \(x = 0.20\,\mathrm{m}\), the electric field due to \(Q_1\) (negative) points left towards the origin, while \(Q_2\) (positive) points right.
  

This determination is essential for calculating net electric field and understanding charge interactions.