Question

A \(100-n C\) point charge is located at \(A(-1,1,3)\) in free space. \((a)\) Find the locus of all points \(P(x, y, z)\) at which \(E_{x}=500 \mathrm{~V} / \mathrm{m} \cdot(b)\) Find \(y_{1}\) if \(P\left(-2, y_{1}, 3\right)\) lies on that locus.

Short answer

Question: Find the locus of all points at which the x-component of the electric field is equal to 500 V/m and a 100-nC point charge is located at point A(-1,1,3). Calculate the value of \(y_1\) for the point P(-2, \(y_1\), 3) lying on that locus. Answer: The locus is given by the equation \((x + 1)^2 + (y - 1)^2 + (z - 3)^2 = \dfrac{50000}{8.99}\). The value of \(y_1\) for the point P(-2, \(y_1\), 3) lying on the locus is \(y_1 = 1 \pm\sqrt{\dfrac{50000}{8.99} - 1}\).

Step-by-step solution

Formula for electric field

We can use the formula for the electric field produced by a point charge: \(E = \dfrac{k \cdot q}{r^3}\vec{r}\) where \(E\) is the electric field vector, \(k = 8.99 \times 10^9 \frac{\mathrm{N m^2}}{\mathrm{C^2}}\) is the electrostatic constant, \(q\) is the charge of the point source, \(r\) is the distance between the point source and the observation point, and \(\vec{r}\) is the position vector.

Compute the position vector and distance

The position vector is the difference between the coordinates of point P and point A: \(\vec{r} = (x - (-1), y - 1, z - 3) = (x + 1, y - 1, z - 3)\) Now we calculate the distance between point A and P: \(r = \sqrt{(x + 1)^2 + (y - 1)^2 + (z - 3)^2}\)

Compute the x-component of the electric field

Substitute the position vector and distance into the electric field formula and take the x-component: \(E_x = \dfrac{k \cdot q}{r^3} \cdot (x + 1)\) Plug in the given charge: \(q = 100 \times 10^{-9}\) C, and set the \(E_x\) value to 500 V/m: \(500 = \dfrac{8.99\times 10^9 \times 100 \times 10^{-9}}{[(x + 1)^2 + (y - 1)^2 + (z - 3)^2]^{3/2}} \cdot (x + 1)\)

Simplify equation and find the locus

Simplify the equation by multiplying both sides by the denominator: \(500[(x + 1)^2 + (y - 1)^2 + (z - 3)^2]^{3/2} = 8.99\times 10^9 \times 100 \times 10^{-9} \cdot (x + 1)\) Divide both sides by \(8.99 \times 10^9\): \(\dfrac{500}{8.99 \times 10^9} [(x + 1)^2 + (y - 1)^2 + (z - 3)^2]^{3/2} = 100 \times 10^{-9} \cdot (x + 1)\) Solve for \((x + 1)^2 + (y - 1)^2 + (z - 3)^2\): \((x + 1)^2 + (y - 1)^2 + (z - 3)^2 = \dfrac{500}{8.99 \times 10^9} \cdot 100 \times 10^9 = \dfrac{50000}{8.99}\)

Find the value of \(y_1\)

Plug in \(x = -2\) and \(z = 3\) into the locus equation: \((-2 + 1)^2 + (y_1 - 1)^2 + (3 - 3)^2 = \dfrac{50000}{8.99}\) Solve for \(y_1\): \((y_1 - 1)^2 = \dfrac{50000}{8.99} - 1^2\) Take the square root of both sides: \(y_1 - 1 = \pm\sqrt{\dfrac{50000}{8.99} - 1}\) Add 1 to both sides: \(y_1 = 1 \pm\sqrt{\dfrac{50000}{8.99} - 1}\) Now, we have the value of \(y_1\) for the point P(-2,\(y_1\),3) lying on the locus.

Key concepts

Understanding the Electric Field Vector

The concept of the electric field vector is central to visualizing and quantifying the influence a charge, specifically a point charge, exerts on its surroundings. An electric field represents the force per unit charge at any point in space due to the presence of charge. In mathematical terms, the electric field vector, denoted as \(E\), at a distance from a point charge is given by the equation \(E = \frac{k \cdot q}{r^3}\vec{r}\), where \(\vec{r}\) is the position vector from the charge to the point of interest.

For students aiming to fully grasp the concept, it's imperative to understand that this vector has both magnitude and direction. The direction of the electric field vector is radial from the point charge and the magnitude diminishes with the cube of the distance from the charge. By focusing on the x-component, \(E_x\), as in the given exercise, we concentrate on the influence along the x-axis which is crucial when examining the effect in a particular direction or dimension.

The Role of the Electrostatic Constant

The electrostatic constant, represented by \(k\), is a proportionality factor that forms part of Coulomb's law, appearing in the electric field equation as \(k = 8.99 \times 10^9 \frac{\mathrm{N m^2}}{\mathrm{C^2}}}\). This constant is pivotal in determining the strength of the electric field generated by a given charge. It quantifies the force per unit charge at a unit distance from the charge.

For students, appreciating the scale of \(k\) helps understand the sensitivity of the electric force to charge and distance. Notably, the higher the value of \(k\), the stronger the electric field produced by a point charge. This fact can be useful when comparing the electric fields resulting from different charges or at different points in space.

Determining the Point Charge Locus

The point charge locus is essentially the collection of points in space where the electric field vector fulfills a specific condition. In the exercise, we are dealing with finding the locus where the x-component of the electric field strength \(E_x\) is exactly 500 V/m. The process is to set up an equation representing the electric field at any point and solve for the coordinates that meet the \(E_x\) value given.

To determine this locus, one must employ the electric field formula and algebraic manipulation. By substituting known values and computing the specific coordinate components that satisfy the equation, the result will yield the locus of points in space that corresponds to the required field condition. In educational practice, this not only teaches students to apply formulas but also reinforces concepts of spatial reasoning and depth perception in electric field analysis.