Calculating Potential from the Field

Calculating Electric Potential from the Electric Field

The relationship between the electric field $\mathbf{E}$ and the electric potential $V$ is fundamental in electromagnetism. The electric potential is related to the electric field through the concept of gradient (or spatial rate of change). Essentially, the electric field is the negative gradient of the electric potential.

1. Electric Field and Electric Potential Relationship

The electric field $\mathbf{E}$ is related to the electric potential $V$ by:

$$\mathbf{E} = -\nabla V$$

In one dimension, this simplifies to:

$$E = -\frac{dV}{dx}$$

Where:

• $\mathbf{E}$ is the electric field (vector),
• $V$ is the electric potential (scalar),
• $\nabla V$ is the gradient (rate of change of potential with respect to position),
• $x$ is the spatial coordinate (in one dimension).

In other words, the electric field is the rate of change of electric potential with respect to position, and it points in the direction of greatest decrease in potential.

2. Calculating Potential from Electric Field in One Dimension

If the electric field $E$ is given as a function of position $x$, you can calculate the electric potential $V$ by integrating the electric field with respect to position:

$$V(x) = - \int E(x) \, dx + C$$

Where:

• $V(x)$ is the electric potential as a function of position $x$,
• $E(x)$ is the electric field as a function of position,
• $C$ is the constant of integration, which can be determined using boundary conditions (such as the value of $V$ at a reference point).

The negative sign arises because the electric field points from higher to lower potential.

3. Example: Calculating Electric Potential from a Constant Electric Field

Consider a constant electric field $E = 500 \, \text{N/C}$ directed along the positive $x$-axis. The potential at $x = 0$ is given as $V(0) = 0$.

To find the potential at a point $x = 2 \, \text{m}$, we use the relationship:

$$V(x) = - \int_0^x E \, dx$$

Since the electric field is constant:

$$V(x) = - E \cdot x + C$$

Substituting $E = 500 \, \text{N/C}$ and $V(0) = 0$, we find $C = 0$. Thus, the potential at $x = 2 \, \text{m}$ is:

$$V(2) = - 500 \cdot 2 = -1000 \, \text{V}$$

So, the electric potential at $x = 2 \, \text{m}$ is $-1000 \, \text{V}$.

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4. Example: Electric Potential from a Varying Electric Field

Now consider an electric field that varies with position. Suppose the electric field is given by:

$$E(x) = 3x^2 \, \text{N/C}$$

Where $E(x)$ is the electric field as a function of position $x$. To find the electric potential at position $x$, integrate $E(x)$:

$$V(x) = - \int_0^x E(x') \, dx'$$

Substitute $E(x') = 3x'^2$:

$$V(x) = - \int_0^x 3x'^2 \, dx'$$

Performing the integration:

$$V(x) = - \left[ x'^3 \right]_0^x = -x^3$$

Thus, the electric potential as a function of position is:

$$V(x) = -x^3 + C$$

To determine $C$, we need a boundary condition (e.g., the potential at $x = 0$ is zero, so $V(0) = 0$):

$$V(0) = - 0^3 + C = 0 \quad \Rightarrow \quad C = 0$$

Therefore, the potential is:

$$V(x) = -x^3$$

This gives the electric potential at any point $x$ in the field.

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5. Electric Potential Due to a Uniform Electric Field

For a uniform electric field $E$, the electric potential difference $\Delta V$ between two points separated by a distance $d$ is given by:

$$\Delta V = - E \cdot d$$

This equation arises from the fact that the electric field is constant and points from higher to lower potential, and the potential difference depends on the distance along the field direction.

For example, if $E = 1000 \, \text{N/C}$ and the points are separated by $5 \, \text{m}$, the potential difference is:

$$\Delta V = - 1000 \times 5 = -5000 \, \text{V}$$

Thus, the potential decreases by 5000 V as you move in the direction of the electric field.

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6. Example: Potential Due to a Point Charge

The electric field $\mathbf{E}$ due to a point charge $Q$ at a distance $r$ is given by:

$$E(r) = \frac{k_e Q}{r^2}$$

To find the potential at a distance $r$ from the charge, integrate the electric field:

$$V(r) = - \int_{\infty}^r E(r') \, dr' = - \int_{\infty}^r \frac{k_e Q}{r'^2} \, dr'$$

Performing the integration:

$$V(r) = - \left[ \frac{k_e Q}{r'} \right]_{\infty}^r = \frac{k_e Q}{r}$$

So, the electric potential at a distance $r$ from a point charge $Q$ is:

$$V(r) = \frac{k_e Q}{r}$$

This is the same result as we would obtain directly from the formula for the electric potential due to a point charge.

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7. Calculating Potential from Field in Higher Dimensions

In higher dimensions, the process of calculating the potential from the electric field follows the same principle of integration, but the electric field will generally depend on the geometry and the coordinate system. For example, in spherical coordinates (for a radially symmetric field, such as the field due to a point charge), the electric field $E(r)$ only depends on $r$, and the potential $V(r)$ is:

$$V(r) = - \int_{\infty}^r E(r') \, dr'$$

For a point charge, the electric field is $E(r) = \frac{k_e Q}{r^2}$, and the potential $V(r)$ is $V(r) = \frac{k_e Q}{r}$, as derived earlier.

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Summary

• The electric field is the negative gradient of the electric potential: $\mathbf{E} = -\nabla V$.
• To calculate the electric potential $V$ from a known electric field $E$, you integrate the electric field: $$V(x) = - \int E(x) \, dx$$
• For a constant electric field, the potential varies linearly with position.
• For a non-uniform electric field, the potential is found by integrating the electric field expression.
• The electric potential due to a point charge is $V(r) = \frac{k_e Q}{r}$, which can be derived from integrating the electric field.

These principles are widely used in electrostatics to calculate potentials and understand the behavior of charges in electric fields.