Question

Three charges are on the \(y\) -axis. Two of the charges, each \(-q\), are located \(y=\pm d\), and the third charge, \(+2 q\), is located at \(y=0\). Derive an expression for the electric field at a point \(P\) on the \(x\) -axis.

Short answer

Answer: The expression for the electric field at point P on the x-axis is \(E_P = 2kq \Bigl(\frac{1}{x^2 + d^2} + \frac{1}{x^2}\Bigr)\).

Step-by-step solution

Defining the Problem and Known Quantities

We have three charges given at the following positions on the y-axis: - Charge -q at y = -d - Charge -q at y = d - Charge +2q at y = 0 We are supposed to find an expression for the electric field at a point P on the x-axis.

Calculate the Electric Field Due to -q at y = -d

Let's denote the distance from charge -q at y = -d to point P on the x-axis as r1. Using the formula for the electric field due to a point charge: \(E_1 = \frac{k\cdot |(-q)|}{r_1^2}\), where k is the Coulomb constant. Now, we need to find an expression for \(r_1\). Since point P is on the x-axis, we can use the Pythagorean theorem with the coordinates (x, 0) for point P and (0, -d) for the charge position. \(r_1 = \sqrt{x^2 + (-d)^2}\) Plug this expression for \(r_1\) into the electric field formula: \(E_1 = \frac{k\cdot q}{(x^2 + d^2)}\)

Calculate the Electric Field Due to -q at y = d

Now, we will repeat the same procedure for the charge -q at y = d. Let's define the distance between the charge and point P as r2. Using the electric field formula: \(E_2 = \frac{k\cdot |(-q)|}{r_2^2}\) We can find the expression for \(r_2\) using the Pythagorean theorem with the coordinates (x, 0) for point P and (0, d) for the charge position: \(r_2 = \sqrt{x^2 + d^2}\) Plug this expression for \(r_2\) into the electric field formula: \(E_2 = \frac{k\cdot q}{(x^2 + d^2)}\)

Calculate the Electric Field Due to +2q at y = 0

In this case, the charge is directly on the x-axis, so the distance between the charge +2q at y = 0 and point P is simply the x-coordinate of point P. Therefore, r3 = x. Using the electric field formula: \(E_3 = \frac{k\cdot (2q)}{r_3^2} = \frac{2kq}{x^2}\)

Calculate the Vector Sum of Electric Fields

Now we need to add up the electric fields due to the three charges. Notice that \(E_1\) and \(E_2\) are equal in magnitude and act along the x-axis. Therefore, these two fields will add up to: \(E_{1+2} = 2E_1 = 2\frac{k\cdot q}{(x^2 + d^2)}\) The total electric field at point P will be the vector sum of \(E_{1+2}\) and \(E_3\). Since \(E_{1+2}\) and \(E_3\) both act along the x-axis, we can easily add them: \(E_P = E_{1+2} + E_3 = 2\frac{k\cdot q}{(x^2 + d^2)} + \frac{2kq}{x^2}\)

Derive the Final Expression for Electric Field at Point P

We can factor out 2kq from the expression which results in: \(E_P = 2kq \Bigl(\frac{1}{x^2 + d^2} + \frac{1}{x^2}\Bigr)\) This is the final expression for the electric field at point P on the x-axis for the given configuration of charges.

Key concepts

Coulomb's Law

Understanding Coulomb's Law is fundamental when studying electric fields and forces between point charges. It states that the magnitude of the electrostatic force (\(F\textsubscript{electrostatic}\textsubscript{})) between two point charges is directly proportional to the product of the magnitudes of the charges (\(q\textsubscript{1}\textsubscript{}\textsubscript{})) and (\(q\textsubscript{2}\textsubscript{}\textsubscript{})), and inversely proportional to the square of the distance (\(r^{2})) between them. The Law is mathematically expressed as:
\[ F = k \frac{\left | q\textsubscript{1}\textsubscript{} q\textsubscript{2}\textsubscript{} \right |}{r^{2}} \] where \(k\) is the Coulomb's constant, approximately equal to \(8.9875 \times 10^{9} N\cdot m^{2}/C^{2}\).When calculating the electric field (\(E\textsubscript{}})) produced by a point charge, Coulomb's Law is adjusted to express the field strength at a point in space, relative to the charge creating the field:
\[ E = k \frac{\left | q \right |}{r^{2}} \] Coulomb's Law is vectorial, meaning it accounts for the direction of the force. When multiple charges are present, their individual fields add up vectorially, which becomes significant in more complex configurations, such as the one presented in our exercise.

Electric Field Due to a Point Charge

The electric field due to a point charge is a vector quantity that represents the force per unit charge exerted on a positive test charge placed at a certain point in space. The intensity of this field diminishes with the square of the distance from the charge, aligning with the principles of Coulomb's Law.For a single charge, the electric field (\(E\textsubscript{}})) can be calculated using the formula:
\[E = \frac{k \cdot |q|}{r^2}\] Where \(k\) is Coulomb's constant, \(q\) is the charge, and \(r\) is the radial distance from the charge to the point of interest.Electric field vectors always start on positive charges and end on negative charges, describing the direction a positive test charge would move. This concept is key when we analyze the cumulative effect of multiple point charges on the electric field at a point in space, like point P in our example.

Pythagorean Theorem in Electric Fields

The Pythagorean theorem is commonly known in geometry for calculating the length of the sides in a right triangle. Physics also uses this theorem to resolve distances in electric fields, especially when charges are positioned in a Cartesian coordinate system.In our exercise, the Pythagorean theorem helps to deduce the distances (\(r_1\) and \(r_2\)) from the charges situated on the \(y\)-axis to a point P on the \(x\)-axis. The formula:\[ r = \sqrt{x^2 + y^2} \] is applied where \(x\) and \(y\) are the distances along the respective axes. These distances are critical when computing the electric field due to point charges, as the field strength is inversely proportional to the square of the distance from the charge. Correctly finding the distance using the Pythagorean theorem is vital for calculating the accurate value of the electric field.

Vector Sum of Electric Fields

When dealing with multiple electric fields generated by various point charges in a system, the total electric field at a given point is the vector sum of the individual fields. Since the electric field is a vector, both its magnitude and direction must be considered during summation.In vector addition, components that lie along the same axis can be simply added or subtracted, according to their direction (positive or negative axis orientation). In our example, the electric fields due to the charges at \(y = \pm d\) point in the same direction along the \(x\)-axis, making their addition straightforward. Conversely, if electric field vectors do not align on the same axis, we would have to break them into their components and use vector addition to determine the resultant field.For example, to calculate the electric field at point P, we found \(E_{1+2}\), the sum of fields due to charges at \(y = \pm d\), and added it to \(E_3\), the field due to the charge at \(y = 0\). Since these fields act along the same direction (the \(x\)-axis), we summed their magnitudes to find the total electric field at P.