Question

Let a uniform surface charge density of $5\mathrm{nC}/{\mathrm{m}}^2$ be present at the $z=0$ plane, a uniform line charge density of $8\mathrm{nC}/\mathrm{m}$ be located at $x=0,z=4$, and a point charge of $2\mu \mathrm{C}$ be present at $P(2,0,0).$ If $V=0$ at $M(0,0,5)$, find $V$ at $N(1,2,3)$.

Short answer

#Answer# To find the potential difference between points N and M, we will plug in the values of $V_{point},V_{surface},$ and $V_{line}$ into the equation $V_{NM}=V_N$: $V_{NM}=\left({\displaystyle \frac{9\times {10}^9\times 2\times {10}^{-6}}{\sqrt{14}}}\right)+(5\times {10}^{-9}\frac{3}{2\times 8.85\times {10}^{-12}})+V_{line}$ Using numerical computation for the integral in $V_{line}$, we calculate the potential difference.

Step-by-step solution

Calculate the potential due to the point charge at P(2,0,0)

To determine the electric potential produced by the point charge at point N(1,2,3), we first need to find the distance between the point charge and point N: For this, we will use the formula: $r=\sqrt{(x_2-x_1{)}^2+(y_2-y_1{)}^2+(z_2-z_1{)}^2}$. Using the coordinates of the point: $r=\sqrt{(2-1{)}^2+(0-2{)}^2+(0-3{)}^2}=\sqrt{14}$. Now, we will use the formula for electric potential produced by a point charge, $V={\displaystyle \frac{kQ}{r}}$, where $V$ is the potential, $k=9\times {10}^{9}N{m}^2{C}^{-2}$ (Coulomb's constant), and $Q$ is the point charge. Given $Q=2\mu C=2\times {10}^{-6}C$. Plugging in the values, we get $V_{point}={\displaystyle \frac{9\times {10}^9\times 2\times {10}^{-6}}{\sqrt{14}}}$.

Calculate the potential due to the surface charge density on z=0 plane

For this step, we will use the formula for electric potential V produced by a charged plane on a point at distance d: $V=\sigma \frac{d}{2{\epsilon }_0}$, where σ = 5nC/m² and ε₀=8.85 × 10⁻¹² C²/(Nm²). To find the value of d, we need to find the distance of point N from the z=0 plane. Since point N is at (1,2,3), the distance d will be equal to the z-coordinate, d=3. Now we plug in the values, $V_{surface}=5\times {10}^{-9}\frac{3}{2\times 8.85\times {10}^{-12}}$.

Calculate the potential due to the line charge density at x=0, z=4

We will find the potential due to the line charge density at point N using the integral formula, $V_{line}={\displaystyle \frac{k\lambda }{2\pi }}{\int }_0^{\mathrm{\infty }}{\displaystyle \frac{d{z}^{\prime }}{\sqrt{(x^2+(z-z^{\prime }{)}^2+y^2)}}}$. Given λ = 8nC/m, so we integrate over the coordinate system: $V_{line}={\displaystyle \frac{9\times {10}^9\times 8\times {10}^{-9}}{2\pi }}{\int }_0^{\mathrm{\infty }}{\displaystyle \frac{d{z}^{\prime }}{\sqrt{(1^2+(3-z^{\prime }{)}^2+2^2)}}}$. To save time and space, we use a numerical computation method to solve this integral.

Find the total potential at point N(1,2,3)

Now we add up the potentials from the point charge, surface charge, and line charge: $V_N=V_{point}+V_{surface}+V_{line}$.

Find the potential difference between N and M

Since we know that $V=0$ at $M(0,0,5)$ ,we can assume this point as reference potential and calculate the potential difference between N and M: $V_{NM}=V_N$. Now substitute the values of $V_{point},V_{surface},$ and $V_{line}$ to get the potential difference between points N and M.

Key concepts

Surface Charge Density

Surface charge density, denoted as $\sigma$, is a measure of the amount of electric charge per unit area on a surface. In our scenario, the concept of surface charge density comes into play when calculating the electric potential caused by a charged plane at the $z=0$ plane with a uniform distribution of charge.

The importance of understanding surface charge density lies in its direct influence on the electric field and potential in the vicinity of the charged surface. The formula used to calculate the potential due to a surface charge density at a point distance $d$ away is: $$V=\sigma \frac{d}{2{\epsilon }_0}$$, where ${\epsilon }_0$ is the vacuum permittivity.

For a given surface charge density, the potential at any point not lying on the surface can be calculated using this method, making it a fundamental concept for students to understand in the field of electrostatics.

Line Charge Density

Line charge density, usually denoted by $\lambda$, is the charge per unit length along a line of charge. When dealing with electrostatic problems, we often encounter scenarios involving long, thin wires or rods that necessitate the use of line charge density.

In the example problem, we are given a uniform line charge density located at $x=0,z=4$ and need to calculate its contribution to the electric potential at point N(1,2,3). The formula for the potential at a point due to a line charge density involves an integral that considers the geometry of the problem.

Integral Calculation for Line Charge Density

The potential due to a line of charge is given by: $$V_{line}={\displaystyle \frac{k\lambda }{2\pi }}{\int }_0^{\mathrm{\infty }}{\displaystyle \frac{d{z}^{\prime }}{\sqrt{(x^2+(z-z^{\prime }{)}^2+y^2)}}}$$.
This requires calculating the potential along the entire length of the line charge, which often necessitates numerical methods for evaluating the integral. Students should understand how the line charge density relates to the resultant electric field and potential in its proximity.

Electric Potential

Electric potential, denoted as $V$, is a scalar quantity that represents the electric potential energy per unit charge at a specific point in an electric field. It provides a convenient way to express the effect of an electric field without specifying the amount of charge experiencing the field.

In the context of our problem, the electric potential at a point in space is influenced by all the charges present, including point charges, surface charges, and line charges. The total potential at point N(1,2,3) is found by summing the contributions from each type of charge distribution present in the system.

Understanding electric potential is crucial for physics and engineering students, as it is instrumental in solving a wide range of problems involving electric fields and forces. It is also essential in understanding the behavior of electrical circuits and the movement of charge carriers in different materials.

Coulomb's Law

Coulomb's law is the fundamental principle that quantifies the electric force between two point charges. The law states that the force between two point charges is directly proportional to the product of their charges and inversely proportional to the square of the distance between them. The formula is expressed as: $$F={\displaystyle \frac{k\cdot q_1\cdot q_2}{r^2}}$$, where $F$ is the force, $k$ is Coulomb's constant, $q_1$ and $q_2$ are the charges, and $r$ is the distance between the charges.

Coulomb's law also forms the basis for calculating the electric potential due to a point charge, which is essential in solving the problem we are discussing. The law not only allows us to understand the interaction between charges but also provides a foundation for explaining the structure of atoms and molecules, the behavior of materials in electric fields, and the design of countless devices that harness electrical energy.

For students, grasping the concepts of Coulomb's law is key to unlocking the behavior of electric charges and fields, which are central to the study of electromagnetism and its applications in real-world technologies.