Problem 43 - University Physics with Modern Physics

Question

Two positive point charges \(q\) are placed on the \(x\) -axis, one al \(x=a\) and one at \(x=-a\) . (a) Find the magnitude and direction of the electric field at \(x=0 .\) (b) Derive an expression for the electric field at points on the \(x\) -axis. Use your result to graph the \(x\) -component of the electric field as a function of \(x,\) for values of \(x\) between \(-4 a\) and \(+4 a .\)

Step-by-Step Solution

Answer

The electric field is zero at \( x=0 \). For \( |x| \neq a \), use \( E(x) = \frac{kq}{(x-a)^2} - \frac{kq}{(x+a)^2} \).

Step 1: Analyze the placement of charges

Two identical positive charges, each with a charge of \( q \), are placed symmetrically on the \( x \)-axis at positions \( x = a \) and \( x = -a \). The setup is symmetric around the origin.

Step 2: Calculate the electric field at x=0

The electric field due to a point charge \( q \) at a distance \( r \) is given by \( E = \frac{kq}{r^2} \), where \( k \) is Coulomb's constant. At \( x=0 \), the distance from both charges is \( a \). Hence, the electric field due to each charge has magnitude \( E = \frac{kq}{a^2} \). Since both charges are positive and equidistant, their fields at the origin are equal and opposite (because they point away from their respective charges), cancelling each other out. Thus, **the net electric field at \( x=0 \) is 0**.

Step 3: Determine the electric field at a general point on the x-axis

Consider a general point \( x \) on the \( x \)-axis. The distance from this point to the charge at \( x = a \) is \(|x - a|\) and to the charge at \( x = -a \) is \(|x + a|\). The electric field from the charge at \( x = a \) will point away from this charge, and similarly for the charge at \( x = -a \). The total electric field is the vector sum of the fields due to each charge.The electric field at position \( x \) from a charge at \( x = a \) is:\[ E_1 = \frac{kq}{(x-a)^2}\text{ (directed towards the left if } x > a, \text{ else towards right)} \]The electric field at position \( x \) from a charge at \( x = -a \) is:\[ E_2 = \frac{kq}{(x+a)^2}\text{ (directed towards the right if } x > -a, \text{ else towards left)} \]The net electric field is the sum of these components:\[ E = E_1 + E_2 = \frac{kq}{(x-a)^2} - \frac{kq}{(x+a)^2} \]

Step 4: Graph the electric field

To graph the \( x \)-component of the electric field, plot \( E(x) = \frac{kq}{(x-a)^2} - \frac{kq}{(x+a)^2} \) for \( x \) ranging from \( -4a \) to \( 4a \). Note the symmetry: the graph will be negative for \( -a < x < a \) and positive for \( |x| > a \). There will be asymptotes at \( x = a \) and \( x = -a \), where the field becomes undefined due to division by zero.

Key Concepts

Point Charges

Point charges are fundamental concepts in the study of electric fields. These are hypothetical charges located at a single point in space, which help us simplify complicated charge distribution calculations. In our scenario, two identical positive point charges are crucial because their symmetric placement around the origin means their effects can combine or cancel depending on the position along the axis.

This symmetry simplifies understanding the electric field at various points. At the origin, for instance, the fields due to each charge are identical in magnitude but opposite in direction, resulting in a net electric field of zero. On the other hand, away from the origin, each charge contributes a field that decreases in intensity with distance according to the inverse square law. This symmetric feature of point charges provides a clear picture of how electric fields operate in multi-charge systems.

By considering these point charges at specific locations, students can develop an intuitive understanding of how electric fields from multiple sources combine through vector addition.

Coulomb's Law

Coulomb's Law is the cornerstone of electrostatics, explaining how point charges exert forces on each other. It tells us that the electric force between these charges is directly proportional to the product of the charges and inversely proportional to the square of the distance between them.

In mathematical terms, it is given by:

• \( F = \frac{k|q_1q_2|}{r^2} \)

Here, \( k \) is a constant known as Coulomb's constant, \( q_1 \) and \( q_2 \) are the magnitudes of the charges, and \( r \) is the distance between them.

Applying this to our problem, we see how the distance from each charge affects the strength of the electric field observed at any given point on the axis. By calculating the fields due to each charge, then finding their sum or difference, we can determine the net electric field as required.

This law provides the essential framework for predicting how these point charges interact and influence the surrounding space through their electric fields.

Electric Field Graphing

Graphing the electric field provides a visual representation of how the field varies over a range of positions. In our example, the electric field's dependence on distance is key to understanding how the point charges affect different regions on the axis.

The function:

• \( E(x) = \frac{kq}{(x-a)^2} - \frac{kq}{(x+a)^2} \)

illustrates how the field changes. For instance, when the graph "dips," it indicates that the electric field points in one direction, usually toward or away from one of the charges.

Plotting this function over a range, such as \(-4a\) to \(4a\), helps us recognize features like symmetry, which reflects the symmetry of the problem itself. The field changes direction at certain points, creating zeros where the forces balance out. Graphs also highlight the concept of asymptotes at locations where the charges themselves are (\(x = a\) and \(x = -a\)), showing how the electric field strength becomes extremely large as you approach a charge.

This graphical analysis not only aids in understanding the behavior of electric fields but also develops an intuitive grasp of how point charges influence their surroundings.