Electric Fields with Calculus: Videos & Practice Problems

Electric Fields with Calculus extends the point-charge field to continuous charge distributions by splitting charge into infinitesimal pieces \(dq\) and adding their contributions with an integral. For a point charge, the vector form is \(\vec{E}=k\frac{Q}{r^3}\vec{r}\) , equivalent to the usual magnitude form with direction set by the sign of the charge. The displacement vector points from the producing charge to the measurement point, while \(r\) is only its magnitude.

For a charge distribution, the total field is found from \(\vec{E}=\int d\vec{E}\) and often from \(dE=k\frac{dq}{r^2}\) . Geometry and symmetry are essential: components perpendicular to the axis of symmetry cancel, leaving only the axial component to integrate. This leads naturally to using linear charge density \(\lambda\) for lines and surface charge density \(\sigma\) for disks, with \(dq=\lambda\\,dy\) and \(dq=\sigma\\,dA\) .

Electric Fields with Calculus

Electric fields generated by point charges can be described using vector calculus, extending the principles of Coulomb's Law. The electric field 𝐄 due to a point charge Q at a distance r is given by the vector form:

\[\mathbf{E} = k \frac{Q}{r^3} \mathbf{r}\]

Here, k is Coulomb's constant, approximately 8.99 × 10⁹ N·m²/C², r is the scalar distance from the charge to the point of interest, and 𝑟 is the position vector pointing from the charge to that point. It is important to distinguish between the scalar r, the unit vector r̂, and the vector 𝑟. The vector 𝑟 includes both magnitude and direction, often expressed in terms of unit vectors such as î and ĵ in Cartesian coordinates.

The direction of the electric field depends on the sign of the charge: it points away from positive charges and toward negative charges. When dealing with continuous charge distributions rather than point charges, the total electric field is found by integrating the contributions from infinitesimal charge elements dQ. Each element produces a small electric field d𝐄, and the total field is the vector sum of all these contributions:

\[\mathbf{E} = k \int \frac{dQ}{r^3} \mathbf{r}\]

This integral accounts for the geometry and distribution of the charge, allowing calculation of the electric field for complex shapes.

For example, consider a point charge located at the origin and a point P at coordinates (0.04 m, 0.03 m). The position vector 𝑟 from the charge to point P is:

\[\mathbf{r} = 0.04\, \hat{i} + 0.03\, \hat{j} \quad \text{(in meters)}\]

The magnitude of this vector is:

\[r = \sqrt{(0.04)^2 + (0.03)^2} = 0.05\, \text{m}\]

Using the electric field formula, if the charge Q is 3 coulombs, the electric field vector at point P is:

\[\mathbf{E} = 8.99 \times 10^9 \times \frac{3}{(0.05)^3} (0.04\, \hat{i} + 0.03\, \hat{j})\]

Calculating this yields components:

\[E_x = 8.63 \times 10^{12} \, \text{N/C} \quad \hat{i} \quad \text{and} \quad E_y = 6.47 \times 10^{12} \, \text{N/C} \quad \hat{j}\]

This vector form of the electric field clearly shows both magnitude and direction, essential for understanding electric forces in two or three dimensions.

Mastering the use of vector calculus in electric fields enables the analysis of more complex charge distributions by integrating infinitesimal contributions, a fundamental skill in electromagnetism and physics.

Electric Fields with Calculus Example 1

When calculating the electric field produced by a ring of charge at a point along its axis, it is essential to apply both geometric reasoning and symmetry principles. The ring is assumed to be uniformly charged, and the goal is to find the electric field at a point P located a distance z from the center of the ring along the axis perpendicular to its plane.

Using the Pythagorean theorem, the distance from any infinitesimal charge element dQ on the ring to the point P is given by the hypotenuse of a right triangle formed by the ring's radius R and the axial distance z. This distance r is expressed as:

\[r = \sqrt{R^2 + z^2}\]

Each infinitesimal charge element dQ produces a small electric field dE at point P, whose magnitude is determined by Coulomb's law:

\[dE = \frac{k \, dQ}{r^2}\]

However, because the ring is symmetric, the horizontal components of the electric field from opposite elements cancel out. Only the vertical components along the axis (the z-direction) add constructively. The vertical component of the infinitesimal electric field is found by multiplying dE by the cosine of the angle θ between the line connecting dQ and point P and the axis:

\[dE_z = dE \cos \theta\]

From the geometry of the triangle, the cosine of the angle θ is the adjacent side over the hypotenuse:

\[\cos \theta = \frac{z}{\sqrt{R^2 + z^2}}\]

Substituting this into the expression for dE_z and combining terms yields:

\[dE_z = \frac{k \, dQ}{R^2 + z^2} \cdot \frac{z}{\sqrt{R^2 + z^2}} = \frac{k z \, dQ}{(R^2 + z^2)^{3/2}}\]

Since the radius R and the axial distance z are constants for all charge elements on the ring, they can be factored out of the integral. Integrating over the entire ring, the total electric field along the axis is:

\[E_z = \frac{k z}{(R^2 + z^2)^{3/2}} \int dQ\]

The integral of dQ over the ring is simply the total charge Q. Therefore, the electric field at point P along the axis of the ring is:

\[E_z = \frac{k Q z}{(R^2 + z^2)^{3/2}}\]

This electric field points along the axis of the ring, either toward or away from the ring depending on the sign of the charge. This result highlights the importance of symmetry in simplifying electric field calculations and demonstrates how integrating infinitesimal contributions leads to the total field. It also emphasizes the use of geometric relationships and the Pythagorean theorem in relating distances and angles in electrostatics problems.

Electric Fields with Calculus Example 2

To calculate the electric field due to a uniformly charged disc at a point along the axis perpendicular to its surface, we consider the disc as composed of many concentric rings of charge. Each ring has a radius r and contributes an infinitesimal electric field component at the point of interest, located a distance z along the axis.

The electric field on the axis due to a ring of charge is given by the formula:

\[E_z = \frac{1}{4 \pi \varepsilon_0} \frac{Q z}{(R^2 + z^2)^{3/2}}\]

where Q is the total charge on the ring, R is the ring radius, and z is the distance along the axis. To find the total field from the disc, we integrate the contributions from rings of radius ranging from 0 to the disc radius R.

Since the charge is uniformly distributed over the disc surface, the surface charge density σ is defined as the total charge divided by the disc area:

\[\sigma = \frac{Q}{\pi R^2}\]

An infinitesimal ring of radius r and thickness dr has an area:

\[dA = 2 \pi r \, dr\]

and thus an infinitesimal charge:

\[dQ = \sigma \, dA = \sigma \cdot 2 \pi r \, dr\]

The infinitesimal electric field contribution from this ring at point P is:

\[dE_z = \frac{1}{4 \pi \varepsilon_0} \frac{dQ \cdot z}{(r^2 + z^2)^{3/2}} = \frac{1}{4 \pi \varepsilon_0} \frac{\sigma \cdot 2 \pi r \, dr \cdot z}{(r^2 + z^2)^{3/2}}\]

Extracting constants and setting up the integral over r from 0 to R:

\[E_z = \frac{\sigma z}{2 \varepsilon_0} \int_0^R \frac{r \, dr}{(r^2 + z^2)^{3/2}}\]

Using the substitution u = r^2 + z^2, so that du = 2r \, dr, the integral becomes:

\[E_z = \frac{\sigma z}{2 \varepsilon_0} \int_{z^2}^{R^2 + z^2} u^{-3/2} \frac{du}{2} = \frac{\sigma z}{4 \varepsilon_0} \int_{z^2}^{R^2 + z^2} u^{-3/2} du\]

Evaluating the integral:

\[\int u^{-3/2} du = -2 u^{-1/2} + C\]

Applying the limits:

\[E_z = \frac{\sigma z}{4 \varepsilon_0} \left[-2 \left(\frac{1}{\sqrt{R^2 + z^2}} - \frac{1}{z}\right)\right] = \frac{\sigma z}{2 \varepsilon_0} \left(\frac{1}{z} - \frac{1}{\sqrt{R^2 + z^2}}\right)\]

Simplifying by canceling z:

\[E_z = \frac{\sigma}{2 \varepsilon_0} \left(1 - \frac{z}{\sqrt{R^2 + z^2}}\right)\]

This expression represents the electric field at a distance z along the axis of a uniformly charged disc of radius R and surface charge density σ.

In the limiting case where the disc radius R is much larger than the distance z (i.e., R/z → ∞), which corresponds to either a very large disc or a point very close to the disc surface, the term:

\[\frac{z}{\sqrt{R^2 + z^2}} \to 0\]

and the electric field approaches the constant value:

\[E_z = \frac{\sigma}{2 \varepsilon_0}\]

This is the well-known result for the electric field near an infinite charged plane, demonstrating that the field is uniform and independent of distance from the plane in this idealized scenario.

Understanding this derivation highlights the importance of surface charge density in two-dimensional charge distributions and the use of integration to sum contributions from infinitesimal charge elements. The approach also illustrates how symmetry and substitution techniques simplify complex electrostatic problems, providing foundational insight into electric fields generated by charged surfaces.

Electric Fields with Calculus Example 3

To determine the electric field generated by a uniformly charged line segment, we start by considering the contributions from infinitesimal charge elements along the line. Each small charge element, denoted as dQ, produces a tiny electric field dE at a point located a distance z from the midpoint of the line on the axis perpendicular to it.

The magnitude of this infinitesimal electric field is given by Coulomb’s law as:

\[dE = \frac{k \, dQ}{r^2}\]

where k is Coulomb’s constant and r is the distance from the charge element to the point of interest. Using the Pythagorean theorem, this distance can be expressed as:

\[r = \sqrt{z^2 + y^2}\]

where y is the position of the charge element along the line segment.

Because the electric field is a vector, we must consider its direction. The vertical components of the electric field from symmetrically opposite charge elements cancel out due to symmetry, leaving only the horizontal components along the z-axis. The horizontal component of the infinitesimal field is:

\[dE_z = dE \cos \theta\]

where the angle θ between the line connecting the charge element to the point and the z-axis satisfies:

\[\cos \theta = \frac{z}{\sqrt{z^2 + y^2}}\]

Substituting these expressions, the infinitesimal electric field component along the z-axis becomes:

\[dE_z = \frac{k \, dQ}{z^2 + y^2} \cdot \frac{z}{\sqrt{z^2 + y^2}} = \frac{k z \, dQ}{(z^2 + y^2)^{3/2}}\]

To integrate over the entire line segment, we relate the infinitesimal charge dQ to the linear charge density λ and the infinitesimal length element dy:

\[\lambda = \frac{Q}{2A} \quad \Rightarrow \quad dQ = \lambda \, dy\]

where Q is the total charge distributed uniformly over a line segment of length 2A. The total electric field along the z-axis is then:

\[E_z = \int_{-A}^{A} \frac{k z \lambda \, dy}{(z^2 + y^2)^{3/2}} = k z \lambda \int_{-A}^{A} \frac{dy}{(z^2 + y^2)^{3/2}}\]

This integral is nontrivial and typically requires advanced techniques such as trigonometric substitution to solve. The evaluated integral yields:

\[\int_{-A}^{A} \frac{dy}{(z^2 + y^2)^{3/2}} = \frac{2A}{z^2 \sqrt{z^2 + A^2}}\]

Substituting back, the electric field becomes:

\[E_z = k z \lambda \cdot \frac{2A}{z^2 \sqrt{z^2 + A^2}} = \frac{2 k A \lambda}{z \sqrt{z^2 + A^2}}\]

Replacing the linear charge density λ with its expression in terms of total charge Q:

\[\lambda = \frac{Q}{2A}\]

we simplify the expression to:

\[E_z = \frac{k Q}{z \sqrt{z^2 + A^2}}\]

This formula gives the magnitude of the electric field at a point along the axis perpendicular to the midpoint of a uniformly charged line segment. The field points along the z-axis, away from the line if the charge is positive. Understanding this derivation highlights the importance of symmetry in vector addition, the use of charge density to relate charge elements to spatial variables, and the application of integral calculus to solve for fields generated by continuous charge distributions.

Here's what students ask on this topic:

How do you calculate the electric field from a continuous charge distribution using calculus?

To calculate the electric field from a continuous charge distribution, you divide the total charge into infinitesimal elements $dq$. Each element produces a small electric field $dE$. The total electric field $E$ at a point is found by integrating these contributions over the entire charge distribution: $E=\int dE$. For each element, the field is given by $dE=k\frac{d}{q}r$, where $r$ is the vector from the charge element to the point of interest and $r$ is its magnitude. Using symmetry and geometry simplifies the integral, often reducing it to one dimension. Charge densities like linear $\lambda$ or surface $\sigma$ help express $d$$q$ in terms of length or area elements, making the integral manageable.

What is the difference between the vector $r$ and the scalar $r$ in electric field calculations?

In electric field calculations, $r$ is the displacement vector pointing from the charge producing the field to the point where the field is measured. It has both magnitude and direction and is often expressed in unit vectors like $i$ and $j$. The scalar $r$ is the magnitude (length) of this vector, representing the distance between the charge and the point. When calculating the electric field vector, the formula uses $r$ divided by $r^3$ to ensure the field has the correct direction and magnitude: $E=k\frac{Q}{r^3}r$. Using the vector form allows you to find components of the field in different directions, which is essential for problems involving multiple dimensions.

How does the sign of the charge affect the direction of the electric field vector?

The sign of the charge determines whether the electric field vector points toward or away from the charge. For a positive charge, the electric field vectors point radially outward, away from the charge, indicating that a positive test charge would be repelled. For a negative charge, the electric field vectors point radially inward, toward the charge, showing that a positive test charge would be attracted. This directionality is embedded in the vector form of the electric field: $E=k\frac{Q}{r^3}r$, where the sign of $Q$ controls the vector's direction along $r$. Understanding this helps in correctly setting up and interpreting electric field problems.

Why do we use integrals when calculating electric fields for charge distributions instead of just Coulomb's law?

Coulomb's law applies directly to point charges, giving the electric field from a single charge. However, many real-world problems involve continuous charge distributions, such as charged rods or surfaces, where charge is spread out rather than concentrated at a point. In these cases, the total electric field is the vector sum of fields from all infinitesimal charge elements. Since there are infinitely many such elements, we use calculus and integrals to sum their contributions precisely. By expressing the charge distribution as $d$$q$ and integrating $E=\int dE$, we account for the geometry and varying distances of each element to the point of interest, which Coulomb's law alone cannot handle.

How do you express the electric field vector in terms of unit vectors for a point charge located at a coordinate?

To express the electric field vector from a point charge at a coordinate, first define the displacement vector $r$ from the charge to the point of interest in terms of unit vectors, typically $i$ and $j$. For example, if the point is 4 cm to the right and 3 cm up from the charge, $r=\; 0.04i+\; 0.03j$ meters. Calculate the magnitude $r\; =\; \backslash ,\; \backslash (\backslash sqrt\{(0.04)^2\; +\; (0.03)^2\}\backslash )\; =\; 0.05\backslash ,\; m$. Then use the vector form of the electric field: $E=k\frac{Q}{r^3}r$. Substitute values and distribute to get components along $i$ and $j$. This method gives the electric field's magnitude and direction in vector form, essential for multidimensional problems.