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

Question: Calculate the electric field at point P on the x-axis due to three charges located on the y-axis: -q at distance d above the origin, -q at distance d below the origin, and +2q at the origin. Answer: The electric field at point P on the x-axis due to the given charges is: \(\vec{E} = \left(\dfrac{-2kqx}{(x^2 + d^2)^{\frac{3}{2}}}+\dfrac{2kq}{x^2}\right)\hat{i}\)

Step-by-step solution

Identify the coordinates of charges and point P

The coordinates of the charges and point P are as follows: Charge 1: (-q) at (0,+d) Charge 2: (-q) at (0,-d) Charge 3: (+2q) at (0,0) Point P: (x,0)

Calculate the electric field due to each charge at point P

We will use the electric field formula for a point charge: \(E=\dfrac{kq}{r^2}\), where k is the Coulomb's constant, q is the charge, and r is the distance from the charge to the point P. For Charge 1: P relative to Charge 1 will be at (x,-d) which makes \(r_1 = \sqrt{x^2 + d^2}\). Hence the electric field due to Charge 1 at Point P will be: \(E_1 = \dfrac{k(-q)}{(x^2 + d^2)}\) For Charge 2: P relative to Charge 2 will be at (x, d) which makes \(r_2 = \sqrt{x^2 + d^2}\). Hence the electric field due to Charge 2 at Point P will be: \(E_2 = \dfrac{k(-q)}{(x^2 + d^2)}\) For Charge 3: Since the Charge 3 is on the y-axis and P is on the x-axis, the distance between them is x. Hence, the electric field due to Charge 3 at Point P will be: \(E_3 = \dfrac{k(+2q)}{x^2}\)

Calculate the components of the electric fields at Point P

Now, we need to find the components of the electric fields at point P in the x and y directions. For Charges 1 and 2, the electric fields will have both x and y components. Let's denote the electric field angles as θ1 and θ2. We find the angles through: \(cos(\theta_1) = \dfrac{x}{r_1}\) and \(cos(\theta_2) = \dfrac{x}{r_2}\) The x-components and y-components of electric fields at P due to Charges 1 and 2 will be: \(E_{1x} = E_1 cos(\theta_1) = \dfrac{k(-q)x}{(x^2 + d^2)^{\frac{3}{2}}}\) \(E_{1y} = E_1 sin(\theta_1) = \dfrac{k(-q)(-d)}{(x^2 + d^2)^{\frac{3}{2}}}\) \(E_{2x} = E_2 cos(\theta_2) = \dfrac{k(-q)x}{(x^2 + d^2)^{\frac{3}{2}}}\) \(E_{2y} = E_2 sin(\theta_2) = \dfrac{k(-q)d}{(x^2 + d^2)^{\frac{3}{2}}}\) Charge 3 produces an electric field only in the x-direction: \(E_{3x} = E_3 = \dfrac{k(+2q)}{x^2}\)

Sum the components of electric fields

According to the principle of superposition, the net electric field at point P will be the vector sum of the electric fields due to all three charges. The net x-component of the electric field at P: \(E_x = E_{1x} + E_{2x} + E_{3x} = \dfrac{k(-q)x}{(x^2 + d^2)^{\frac{3}{2}}}+\dfrac{k(-q)x}{(x^2 + d^2)^{\frac{3}{2}}}+\dfrac{k(+2q)}{x^2} = \dfrac{-2kqx}{(x^2 + d^2)^{\frac{3}{2}}}+\dfrac{2kq}{x^2}\) Since \(E_{1y}\) and \(E_{2y}\) have opposite directions, they will cancel each other out: The net y-component of the electric field at P: \(E_y = E_{1y} + E_{2y} = 0\)

Write the final expression for the electric field at point P

The net electric field at Point P on the x-axis is the vector sum of the x and y components: \(\vec{E} = E_x\hat{i} + E_y\hat{j} = \left(\dfrac{-2kqx}{(x^2 + d^2)^{\frac{3}{2}}}+\dfrac{2kq}{x^2}\right)\hat{i}\) Therefore, the electric field at point P on the x-axis due to the given charges is: \(\vec{E} = \left(\dfrac{-2kqx}{(x^2 + d^2)^{\frac{3}{2}}}+\dfrac{2kq}{x^2}\right)\hat{i}\)

Key concepts

Superposition Principle

The superposition principle is a key concept in physics used to simplify the analysis of systems influenced by multiple sources. In the context of electric fields, this principle states that the net electric field due to multiple charges is the vector sum of the individual fields created by each charge.
This means that if you have several charges exerting an electric field at a point, you calculate the electric field due to each charge independently, then add those fields together to get the total electric field.

• Each electric field due to individual charges is calculated separately.
• The vector sum is taken to determine the resultant field.
• The principle applies regardless of how many charges are present.

This method greatly simplifies the calculations since electric fields are vectors, which means they have both magnitude and direction. Using vector algebra, you combine these values considering both these properties to find a single, resultant vector that represents the total electric field.

Point Charge

In electric field problems, a point charge refers to an idealized model where a charged object is considered to be concentrated at a single point in space. This simplification helps in calculating the electric fields comfortably without worrying about the dimensions of the charge itself.
A point charge can be positive or negative, affecting how it interacts with other charges in its vicinity.

• The electric field due to a point charge is radial, emanating outwards for positive charges and inwards for negative charges.
• The strength of the field decreases with the square of the distance from the charge.
• Point charges are used as the basic building blocks for more complicated electric field calculations.

Using point charges in calculations, engineers and physicists can solve complex problems effectively by focusing on only the essential aspects of charge interactions without additional complexity.

Coulomb's Law

Coulomb's Law is fundamental to understanding electric interactions between charged objects. It gives us the electric force between two point charges. This law states that the force between two charges varies directly with the product of their magnitudes and inversely with the square of the distance between them. Mathematically, it's expressed as:
\( F = \dfrac{k |q_1 q_2|}{r^2} \)
where \( F \) is the force between the charges, \( k \) is the Coulomb's constant, \( q_1 \) and \( q_2 \) are the amounts of the charges, and \( r \) is the distance between them.

• Coulomb's Law helps in determining the direction and magnitude of the force.
• The force is attractive if the charges are of opposite signs and repulsive if they have the same sign.
• This equation forms the basis for calculating electric fields, as the electric field is defined as the force per unit charge.

Understanding Coulomb's Law is crucial not just for calculating specific forces, but also for grasping how charges interact at fundamental levels.