Potential Difference with Calculus: Videos & Practice Problems

Potential Difference with Calculus connects electric potential difference to the electric field through the line integral \( \Delta V=-\int \vec{E}\cdot d\vec{x} \) . This is the calculus form of the familiar relation \( E=-\frac{\Delta V}{\Delta x} \) when the field is constant. The negative sign shows that potential decreases in the direction of the field.

For a point charge, the field changes with distance, so integration is required. Using \( \vec{E}=\frac{kQ}{x^2}\hat{i} \) leads to the standard potential result \( V=\frac{kQ}{r} \) . For a distribution of charges, the charge is split into small pieces \(dq\), each contributing \( dV=\frac{k\\,dq}{r} \) , and these contributions are integrated over the whole distribution.

Potential Difference with Calculus

The electric potential difference, often called the potential difference, can be rigorously defined using calculus. It is the difference in electric potential between two points and is related to the electric field through the integral of the electric field vector dotted with an infinitesimal displacement vector. Mathematically, the potential difference ΔV between two points is expressed as the negative integral of the electric field E dotted with the differential displacement d𝑥:

\[\Delta V = - \int \mathbf{E} \cdot d\mathbf{x}\]

This integral accounts for variations in the electric field along the path between the two points. When the electric field is constant, this relationship simplifies to the familiar formula:

\[E = -\frac{\Delta V}{\Delta x}\]

which shows that the electric field is the negative rate of change of the electric potential with respect to distance.

For point charges, the electric potential at a distance R from a charge Q is given by the equation:

\[V = \frac{kQ}{R}\]

where k is Coulomb's constant, approximately equal to 8.99 × 10⁹ N·m²/C². However, when dealing with a distribution of charges rather than a single point charge, the total potential at a point is found by integrating the contributions from each infinitesimal charge element dQ. Each element contributes a small potential dV given by:

\[dV = \frac{k \, dQ}{r}\]

where r is the distance from the charge element to the point of interest. Summing these contributions through integration yields the total potential due to the charge distribution.

To illustrate the use of calculus in calculating potential difference, consider a point charge Q located at the origin along the x-axis. The goal is to find the potential difference between a point at distance R from the charge and a point infinitely far away. Since the electric field from a point charge varies with distance as:

\[\mathbf{E} = \frac{kQ}{x^2} \hat{i}\]

the potential difference is calculated by integrating the electric field from R to infinity:

\[\Delta V = - \int_{R}^{\infty} \frac{kQ}{x^2} \, dx = -kQ \int_{R}^{\infty} x^{-2} \, dx\]

Evaluating the integral gives:

\[\Delta V = -kQ \left[-\frac{1}{x}\right]_{R}^{\infty} = kQ \left(0 - \frac{1}{R}\right) = -\frac{kQ}{R}\]

This result confirms that the potential at a distance R from a point charge is indeed V = kQ/R, consistent with the known formula. The negative sign indicates the direction of potential decrease as one moves away from the charge.

Understanding the relationship between electric field and potential difference through calculus is essential for analyzing complex charge distributions and varying fields. It highlights how the electric potential is a scalar quantity derived from the vector electric field and how integration plays a crucial role in calculating potentials in non-uniform fields.

Potential Difference with Calculus Example 1

To calculate the electric potential at a point due to a line of charge, consider a linear charge distribution with total charge Q spread over a length from -A to A. The point of interest lies on the z-axis at a distance z from the origin. The key is to treat the line charge as composed of infinitesimally small elements of charge, dQ, each contributing a small potential dV at the point.

The potential contribution from a small charge element dQ located at a height y on the line is given by the formula for a point charge potential:

\[dV = \frac{k \, dQ}{r}\]

where k is Coulomb's constant and r is the distance from the charge element to the point on the z-axis. Using the Pythagorean theorem, this distance is:

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

Since the charge is distributed along the y-axis, the infinitesimal charge element can be expressed in terms of the linear charge density λ, defined as the charge per unit length:

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

Thus, the infinitesimal charge element is:

\[dQ = \lambda \, dy\]

Substituting into the expression for dV gives:

\[dV = \frac{k \lambda \, dy}{\sqrt{y^2 + z^2}}\]

To find the total potential V at the point, integrate dV over the length of the charge distribution from -A to A:

\[V = k \lambda \int_{-A}^{A} \frac{dy}{\sqrt{y^2 + z^2}}\]

This integral evaluates to the natural logarithm function:

\[V = k \lambda \left[ \ln \left( y + \sqrt{y^2 + z^2} \right) \right]_{-A}^{A}\]

Applying the limits yields:

\[V = k \lambda \ln \left( \frac{A + \sqrt{A^2 + z^2}}{-A + \sqrt{A^2 + z^2}} \right)\]

This expression represents the electric potential at a distance z along the axis perpendicular to the line charge. Understanding how to convert a continuous charge distribution into an integral over infinitesimal charge elements and applying the linear charge density is essential for solving electrostatics problems involving distributed charges. The use of logarithmic functions in the solution highlights the relationship between geometry and potential in such configurations.

Potential Difference with Calculus Example 2

To calculate the electric potential due to a ring of charge Q with radius R at a point located a distance z along the axis perpendicular to the plane of the ring, we start by considering the contributions from infinitesimal charge elements on the ring. Each small charge element, denoted as dQ, produces a potential dV at the point of interest.

The electric potential dV due to a point charge is given by the formula:

\[dV = \frac{k \, dQ}{r}\]

where k is Coulomb's constant and r is the distance from the charge element to the point where the potential is being calculated. For a ring of radius R, the distance r to the point on the axis at distance z is found using the Pythagorean theorem:

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

Since every charge element on the ring is the same distance r from the point, the potential contributions from all elements can be summed by integrating over the total charge Q:

\[V = \int dV = \int \frac{k \, dQ}{\sqrt{R^2 + z^2}}\]

Because k and r are constants with respect to the integration variable Q, they can be factored out of the integral:

\[V = \frac{k}{\sqrt{R^2 + z^2}} \int_0^Q dQ = \frac{kQ}{\sqrt{R^2 + z^2}}\]

This result shows that the electric potential at a point along the axis of a uniformly charged ring depends on the total charge Q, the radius of the ring R, and the axial distance z. The potential decreases with increasing distance from the ring, following an inverse relationship with the square root of the sum of the squares of R and z. This formula is fundamental in electrostatics for understanding potentials generated by symmetrical charge distributions.

What is the electric potential due a disk of charge, $Q$, with some radius, $R$, at the point shown in the following figure?
(hint: see "Electric Potential due to Ring of Charge")

• A. $k\sigma \pi R$
• B. $\frac{k\sigma \pi R^2}{\sqrt{R^2+z^2}}$
• C. $\frac{k\sigma \pi R}{2}[\sqrt{R^2+z^2}+z]$
• D. $2k\sigma \pi [\sqrt{R^2+z^2}-z]$

Answer: (D)

Here's what students ask on this topic:

What is the relationship between electric potential difference and electric field using calculus?

The electric potential difference \( \Delta V \) between two points is related to the electric field \( \vec{E} \) through the line integral \( \Delta V = -\int \vec{E} \cdot d\vec{x} \). This means the potential difference is the negative integral of the electric field dotted with an infinitesimal displacement vector \( d\vec{x} \). When the electric field is constant, this reduces to the familiar formula \( E = -\frac{\Delta V}{\Delta x} \), where \( E \) is the magnitude of the electric field and \( \Delta x \) is the distance between points. The negative sign indicates that the potential decreases in the direction of the electric field. This calculus approach generalizes the concept to cases where the electric field varies with position.

How do you calculate the electric potential due to a point charge using integration?

For a point charge \( Q \) located at the origin, the electric field varies with distance \( x \) as \( \vec{E} = \frac{kQ}{x^2} \hat{i} \), where \( k \) is Coulomb's constant. To find the potential difference between a point at distance \( R \) and infinity, we use the integral \( \Delta V = -\int_R^{\infty} \vec{E} \cdot d\vec{x} = -\int_R^{\infty} \frac{kQ}{x^2} dx \). Evaluating this integral gives \( \Delta V = \frac{kQ}{R} \), which matches the known formula for the potential due to a point charge. This method is necessary because the electric field is not constant and changes with distance.

How is the potential difference calculated for a distribution of charges?

For a distribution of charges, the total charge \( Q \) is divided into infinitesimal elements \( dq \), each contributing a small potential \( dV = \frac{k \, dq}{r} \), where \( r \) is the distance from the charge element to the point of interest. To find the total potential or potential difference, you integrate these contributions over the entire charge distribution: \( V = \int \frac{k \, dq}{r} \). This integral sums the effects of all small charges, accounting for their varying distances, and is essential when the charge is spread out rather than concentrated at a point.

Why can't we use the simple formula \( E = -\frac{\Delta V}{\Delta x} \) for point charges?

The formula \( E = -\frac{\Delta V}{\Delta x} \) assumes a constant electric field between two points, which is not true for point charges. The electric field from a point charge varies with distance as \( \frac{1}{x^2} \), so it changes continuously along the path. Because of this variation, the electric field cannot be taken out of the integral, and we must use the integral form \( \Delta V = -\int \vec{E} \cdot d\vec{x} \) to accurately calculate the potential difference. This integral accounts for the changing field strength at each point along the path.

What does the negative sign in the integral \( \Delta V = -\int \vec{E} \cdot d\vec{x} \) signify?

The negative sign in the integral \( \Delta V = -\int \vec{E} \cdot d\vec{x} \) indicates that the electric potential decreases in the direction of the electric field. Physically, this means that moving along the direction of the electric field results in a drop in electric potential. This sign convention aligns with the fact that positive charges naturally move from regions of higher potential to lower potential, following the direction of the electric field.