The electric field E and the electric potential V are related as shown in Equation 25.3. We now show how to calculate the value of the electric field if the electric potential is known in a certain region.
From Equation 25.3 we can express the potential difference dV between two points a distance ds apart as
$dV=-\mathbf{E}.d\mathbf{s}$ (1)
If the electric field has only one component $E_x$, then $\mathbf{E}.d\mathbf{s} = E_x.d_x$. Therefore, Equation 25.15 becomes $dV =-E_x.dx$, or
$E_x=-\frac{dV}{dx}$
That is, the x component of the electric field is equal to the negative of the derivative of the electric potential with respect to x. Similar statements can be made about the y and z components. Equation 25.16 is the mathematical statement of the fact that the electric field is a measure of the rate of change with position of the electric potential, as mentioned in Section 25.1.
Experimentally, electric potential and position can be measured easily with a voltmeter (see Section 28.5) and a meter stick. Consequently, an electric field can be determined by measuring the electric potential at several positions in the field and making a graph of the results. According to Equation 25.16, the slope of a graph of V versus x at a given point provides the magnitude of the electric field at that point.
When a test charge undergoes a displacement $d\mathbf{s}$ along an equipotential surface, then $dV = 0$ because the potential is constant along an equipotential surface. From Equation 25.15, we see that $dV = \mathbf{E}.d\mathbf{s}= 0$; thus, E must be perpendicular to the displacement along the equipotential surface. This shows that the equipotential surfaces must always be perpendicular to the electric field lines passing through them.
As mentioned at the end of Section 25.2, the equipotential surfaces for a uniform electric field consist of a family of planes perpendicular to the field lines. Figure 25.13a shows some representative equipotential surfaces for this situation.
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| Fig.1:Equipotential surfaces (the dashed blue lines are intersections of these surfaces with the page) and electric field lines (red-brown lines) for (a) a uniform electric field produced by an infinite sheet of charge, (b) a point charge, and (c) an electric dipole. In all cases, the equipotential surfaces are perpendicular to the electric field lines at every point. |
If the charge distribution creating an electric field has spherical symmetry such that the volume charge density depends only on the radial distance r, then the electric field is radial. In this case, $\mathbf{E}. d\mathbf{s}= E_r dr$, and we can express dV in the form $dV =- E_r.dr$. Therefore,
$E_r=\frac{dV}{dr}$
For example, the electric potential of a point charge is $V = k_eq/r$. Because V is a function of r only, the potential function has spherical symmetry. Applying Equation 25.17, we find that the electric field due to the point charge is $E_r = k_eq/r^2$, a familiar result. Note that the potential changes only in the radial direction, not in any direction perpendicular to r. Thus, V (like $E_r$) is a function only of r. Again, this is consistent with the idea that equipotential surfaces are perpendicular to field lines. In this case the equipotential surfaces are a family of spheres concentric with the spherically symmetric charge distribution (Fig. 25.13b).
The equipotential surfaces for an electric dipole are sketched in Figure 25.13c.
In general, the electric potential is a function of all three spatial coordinates. If V(r) is given in terms of the Cartesian coordinates, the electric field components $E_x$, $E_y$, and $E_z$ can readily be found from V(x, y, z) as the partial derivatives$^3$.
$E_x=\frac{\partial V}{\partial x}$; $E_y=\frac{\partial V}{\partial y}$; $E_z=\frac{\partial V}{\partial z}$
For example, if $V=3x^2y+y^2+yz$, then
$\frac{\partial V}{\partial x}=\frac{\partial}{\partial x}(3x^2y+y^2+yz)=\frac{\partial}{\partial x}(3x^2y)=3y \frac{d}{dx}(x^2)=6xy$
The electric field is constant throughout the region, with a magnitude of 7.00V/m and a direction toward the negative x-axis (since b < 0 ).
Example 2:
Figure 2 shows several equipotential lines each labeled by its potential in volts. The distance between the lines of the square grid represents 1.00 cm. (a) Is the magnitude of the field larger at A or at B? Why? (b) What is E at B? (c) Represent what the field looks like by drawing at least eight field lines.
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| Fig.2 |
Solution:
(a) The electric field is related to the spacing of equipotential lines. Where the equipotential lines are closer together, the electric field is stronger.
- At A, the equipotential lines are closer together compared to B, where they are more spread out.
- Therefore, the electric field is larger at A because the potential changes more rapidly over a smaller distance.
- At B, the equipotential lines from 2 V to 4 V are separated by
- $\Delta V=4-2=2 V$
- $\Delta r=0.0100 \ m$
- Draw at least 8 lines emanating radially outward from the center (since the equipotential lines are circular, the source of the electric field is at the center).
- Ensure the field lines are denser where the equipotential lines are closer together (near A) and more spread out where the equipotential lines are farther apart (near B).
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