Electric Field from Potential Functions with Calculus: Videos & Practice Problems

Electric Field from Potential Functions with Calculus focuses on finding the electric field when the electric potential V is given as a function of position. The key idea is that each field component comes from a partial derivative of the potential, with a negative sign: \(E_x=-\frac{\partial V}{\partial x},\\; E_y=-\frac{\partial V}{\partial y},\\; E_z=-\frac{\partial V}{\partial z}\) . This connects potential and field in the reverse direction of using integrals to find potential from a known field.

A partial derivative means differentiating with respect to only one variable while treating the others as constants. So when finding \(E_x\), only the x-dependence in V is differentiated; for \(E_y\), only the y-dependence is differentiated. The full electric field can also be written in vector form as \( \vec{E}=-\frac{\partial V}{\partial x}\hat{i}-\frac{\partial V}{\partial y}\hat{j}-\frac{\partial V}{\partial z}\hat{k}\) . If the x and y components are known, the field magnitude follows from \(|\vec{E}|=\sqrt{E_x^2+E_y^2}\) .

Calculating Electric Field as Derivative of Potential

Calculating the electric field from a given electric potential involves using partial derivatives, which differ slightly from ordinary derivatives by focusing on one variable at a time while treating others as constants. When the electric potential V is expressed as a function of spatial variables such as x, y, and z, the components of the electric field can be found by taking the negative partial derivatives of V with respect to each coordinate. Specifically, the x-component of the electric field, E_x, is given by \(E_x = -\frac{\partial V}{\partial x}\), the y-component by \(E_y = -\frac{\partial V}{\partial y}\), and the z-component by \(E_z = -\frac{\partial V}{\partial z}\).

Partial derivatives are calculated by differentiating the function with respect to one variable while treating all other variables as constants. For example, if the potential is given by \(V = 5x^2 y - 8 x y^2\), to find \(E_x\), differentiate with respect to x:

\(E_x = -\frac{\partial}{\partial x}(5x^2 y - 8 x y^2) = -\left(10 x y - 8 y^2\right) = -10 x y + 8 y^2\)

Here, y is treated as a constant during differentiation. Similarly, to find \(E_y\), differentiate with respect to y:

\(E_y = -\frac{\partial}{\partial y}(5x^2 y - 8 x y^2) = -\left(5 x^2 - 16 x y\right) = -5 x^2 + 16 x y\)

Once the components of the electric field are determined, the magnitude of the electric field vector at a specific point can be calculated using the Pythagorean theorem for vector components:

\(|\mathbf{E}| = \sqrt{E_x^2 + E_y^2 + E_z^2}\)

For instance, at coordinates \(x=2\) and \(y=0.4\), substituting into the expressions for \(E_x\) and \(E_y\) yields numerical values for each component. Squaring these, summing, and taking the square root gives the magnitude of the electric field, which is measured in newtons per coulomb (N/C).

In vector notation, the electric field can be succinctly expressed as:

\(\mathbf{E} = -\left(\frac{\partial V}{\partial x} \hat{i} + \frac{\partial V}{\partial y} \hat{j} + \frac{\partial V}{\partial z} \hat{k}\right)\)

This compact form highlights the relationship between the electric field and the spatial rate of change of the electric potential in all three dimensions. Understanding how to compute the electric field from a potential function using partial derivatives is essential for analyzing electrostatic problems involving varying potentials in space.

In a region of space, the electric potential is $V(x,y,z)=4xy-7x^2+3yz$. Calculate the $x$, $y$, and $z$ components of the electric field.

• A. $E_x=-4y+14x,E_y=-4x-3z,E_z=-3x$
• B. $E_x=-4y+14x,E_y=-4x-3z,E_z=-3y$
• C. $E_x=-4y-3z,E_y=-4x+14y,E_z=-3y$
• D. $E_x=-4y-3y,E_y=-4x+14y,E_z=-3x$

Answer: (B)

Here's what students ask on this topic:

How do you calculate the electric field components from a given electric potential function?

To calculate the electric field components from a given electric potential function $V$, you use the relationship between the electric field and the gradient of the potential. Each component of the electric field is the negative partial derivative of the potential with respect to that coordinate. Specifically, the components are given by $E{x}_{}=-\frac{\partial }{\mathrm{\partial x}}V$, $E{y}_{}=-\frac{\partial }{\mathrm{\partial y}}V$, and $E{z}_{}=-\frac{\partial }{\mathrm{\partial z}}V$. This means you take the partial derivative of the potential with respect to each variable while treating the other variables as constants, then apply a negative sign. This process reverses the integral relationship used to find potential from a known electric field.

What is a partial derivative and how is it used in finding the electric field from a potential function?

A partial derivative is the derivative of a function with respect to one variable while treating all other variables as constants. In the context of finding the electric field from a potential function $\mathrm{V(x,y,z)}$, partial derivatives allow you to isolate how the potential changes along each coordinate axis independently. For example, to find the $x$-component of the electric field, you compute the partial derivative of $V$ with respect to $x$, denoted as $\frac{\partial }{\mathrm{\partial x}}V$, while treating $y$ and $z$ as constants. The electric field component is then the negative of this partial derivative: $E{x}_{}=-\frac{\partial }{\mathrm{\partial x}}V$. This method is repeated for the $y$ and $z$ components to find the full electric field vector.

How do you find the magnitude of the electric field from its components?

Once you have the components of the electric field, $E{x}_{}$, $E{y}_{}$, and $E{z}_{}$, you can find the magnitude of the electric field vector using the Pythagorean theorem extended to three dimensions. The magnitude is given by $|\Omega |=\sqrt{E^{x_{}}+E^{y_{}}+E^{z_{}}}$. If the problem is two-dimensional, you simply omit the $z$ component. This calculation gives the total strength of the electric field at a point, combining all directional components.

What is the vector form of the electric field in terms of the electric potential?

The electric field vector $\Omega$ can be expressed in terms of the electric potential $V$ using the gradient operator. The vector form is $\Omega =-\frac{\nabla }{\nabla }V$, which expands to $\Omega =-\frac{\partial }{\mathrm{\partial x}}Vᵢ+-\frac{\partial }{\mathrm{\partial y}}Vᵣ+-\frac{\partial }{\mathrm{\partial z}}Vᵤ$, where $ᵢ$, $ᵣ$, and $ᵤ$ are the unit vectors in the $x$, $y$, and $z$ directions respectively. This compact vector notation summarizes the three component equations and is useful for understanding the direction and magnitude of the electric field derived from the potential.

Can you provide an example of calculating the electric field components from a potential function involving multiple variables?

Consider the electric potential function $V=\; 5x{x}^{2}y-\; 8x\; y{y}^2$. To find the $x$-component of the electric field, calculate the partial derivative of $V$ with respect to $x$: $\frac{\partial }{\mathrm{\partial x}}V=\; 10\; x\; y\; -\; 8\; y{y}^2$. Applying the negative sign, $E{x}_{}=\; -10\; x\; y\; +\; 8\; y{y}^2$. For the $y$-component, differentiate with respect to $y$: $\frac{\partial }{\mathrm{\partial y}}V=\; 5\; x{x}^2-\; 16\; x\; y$, so $E{y}_{}=\; -5\; x{x}^2+\; 16\; x\; y$. This example shows how to treat other variables as constants during partial differentiation and then apply the negative sign to find the electric field components.