Conservative nature of Electric fields

The conservative nature of electrostatic fields establishes the electrostatic field as a conservative field. This opens the door for conservation of energy and the work-kinetic energy theorem in discussions of electric fields and electric forces.

In mechanics, the conversation starts with static discussions such as forces. Quickly, though, the necessity for the study of movement leads to a definition of work. The conservative nature of the electric field allows for quick application of the mechanical concept of work to electric problems concerning work, energy, velocity and displacement.

Recall from mechanics that work is defined as the dot product of the force and displacement vectors, \(W=\vec{F}\bullet\vec{d}\). This begets a few very useful rules in the study of electrostatics.

1) If a particle is displaced perpendicular to the electric field, no work is done on the particle by the field.

2) The work done by the electric field is \(W=qEd_{\parallel}\).

1) \(W=\vec{F}\bullet\vec{d}=q\vec{E}\bullet\vec{d}=qEd\cos(90)=0 \text{J}\)

2) \(W=\vec{F}\bullet\vec{d}=q\vec{E}\bullet\vec{d}=qE(d\cos\theta)=qEd_{\parallel}\)

A particle with charge \(q = -6\mu \text{C}\) moves from (0 \(\text{m}\), 0 \(\text{m}\)) to (1 \(\text{m}\), 2 \(\text{m}\)) in a region with electric field \(\vec{E}=(7.0 \frac{\text{N}}{\text{C}}) \hat{x}\). Find the work done by the electric field.

Solution: Use the theorem that says the work done by the electric field is \(W=qEd_{\parallel}\). Since the electric field points in the x direction, the parallel displacement is just \(d_\parallel=\Delta x = x_f - x_0 = 1 - 0 = 1 \text{m}\).

\(W=qEd_{\parallel}=(-6 \mu \text{C})(7 \frac{\text{N}}{\text{C}})(1 \text{m})\)

\(W = -4.2 \times 10^5 \text{J}\)

Combining the work-kinetic energy theorem, \(W=\Delta K\), and conservation of energy, with the formula above gives

\(qEd_{\parallel}=\Delta K\)

Now that kinetic energy, \(K=\dfrac{1}{2}mv^2\), has been brought into the mix, the velocity of a particle passing through an electric field can quickly be calculated.

A 2.1 \(\text{kg}\) particle with charge \(q = 5\mu \text{C}\) starts from rest then moves from (2 \(\text{m}\), 1 \(\text{m}\)) to (3 \(\text{m}\), 4 \(\text{m}\)) in a region with electric field \(\vec{E}=(7.0 \frac{\text{N}}{\text{C}}) \hat{y}\). Find its final velocity.

Solution:

\(qEd_{\parallel}=\Delta K\)

\(qEd_{\parallel}=K_f - K_0\)

\(qEd_{\parallel}=K_f\)

\(qEd_{\parallel}=\dfrac{1}{2}mv_f^2\)

\(v_f=\sqrt{\dfrac{2qEd_{\parallel}}{m}}\)

Since the electric field points in the y direction, the parallel displacement is just \(d_\parallel=\Delta y = y_f - y_0 = 4 - 1 = 3 \text{m}\).

\(v_f=\sqrt{\dfrac{2qEd_{\parallel}}{m}} = \sqrt{\dfrac{2(5\mu C)(7.0 \dfrac{N}{C})(3 m)}{2.1 kg} } = 10^{-2} \frac{\text{m}}{\text{s}}\)

A conservative field is one that can be written as the gradient of a scalar potential V as

\(\vec{E}=\nabla V\)

The scalar potential of the electric field is known to be \(V=\dfrac{kQ}{r}\).

\(\vec{E}=\nabla V\)

The gradient in spherical coordinates is \(\nabla = {\partial \over \partial r}\hat{r} + {1 \over r}{\partial \over \partial \theta}\hat{\theta} + {1 \over r\sin\theta}{\partial \over \partial \varphi}\hat{\varphi}\), so

\(\nabla V ={\partial V \over \partial r}\hat{r} + {1 \over r}{\partial V \over \partial \theta}\hat{\theta} + {1 \over r\sin\theta}{\partial V \over \partial \varphi}\hat{\varphi}.\)

Evaluate each of the partial derivatives of the scalar potential equation, \(V=\dfrac{kQ}{r}\).

\(\dfrac{\partial V }{\partial r} =- \dfrac{kQ}{r^2}\)

\(\dfrac{\partial V }{\partial \theta} = 0\)

\(\dfrac{\partial V }{\partial \varphi} = 0\)

Using these partial derivatives with the definition for the gradient, only the \(\hat{r}\) component survives, since the other components contain partial derivatives that are equal to 0.

\(\vec{E}={\partial V \over \partial r}\hat{r} + {1 \over r}{\partial V \over \partial \theta}\hat{\theta} + {1 \over r\sin\theta}{\partial V \over \partial \varphi}\hat{\varphi}\)

\(\vec{E}=-\dfrac{kQ}{r^2}\hat{r}\)

Note: The negative sign is the reason that the relationship between the electric field and scalar potential is actually written

\(\vec{E}=-\nabla V\)