# Electrostatic field in a vacuum

Can an electrostatic field, the direction of which is constant in space, but the magnitude varies in space, exist in a vacuum?

• yes! as long as there's something outside the vacuum at least... – danimal Apr 27 '15 at 16:18
• if you put two parallel, oppositely charged plates on either side of a vacuum you would have a constant-direction electric field – danimal Apr 27 '15 at 16:23
• @danimal Sorry i was confused of why you said yes ... Now its clear :P – Shashank Apr 27 '15 at 16:29
• If the direction is constant doesn't that mean the field lines will all be parallel, in which case the magnitude can't vary? – John Rennie Apr 27 '15 at 16:51

No, this is not possible.

Consider a field which always points in the same direction, and put your $z$ axis in that direction. Your field can then be described as $$\mathbf E=E_z(x,y,z)\hat{\mathbf z}.$$ As an electrostatic field, this must satisfy Gauss's law, which in vacuum reads $$\nabla\cdot\mathbf E=\frac{\partial E_z}{\partial z}=0,$$ and means $E_z$ cannot depend on the $z$ coordinate. More intuitively, the electric field cannot change its magnitude along its direction in the absence of electric charge.

In addition, this electrostatic field must be curl-less (or it wouldn't be electrostatic). The $x$ and $y$ components of the equation $\nabla\times\mathbf E=0$ read $$\frac{\partial E_z}{\partial y}=\frac{\partial E_z}{\partial y}-\frac{\partial E_y}{\partial z}=0$$ and $$-\frac{\partial E_z}{\partial x}=\frac{\partial E_x}{\partial z}-\frac{\partial E_z}{\partial x}=0$$ so there can't be any variation along $x$ or $y$ either. More intuitively, if the field magnitude varied along $x$ then a rectangular loop with its edges along $x$ and $z$ would have nonzero circulation.

Taken together, these conditions imply a field in vacuum whose direction is constant must also have constant magnitude.

No. I'm going to assume you mean we can do things to the boundary of a space, but that inside the space has no charge. Let's say $\vec{E}$ is in the constant $z$-direction, and call it $E_z$. Then in vacuum Gauss's Law requires $\nabla \cdot E = \partial_z E_z = 0$. So you can't have a spatial variation in the same direction as this field, but maybe you could make an $E_z$ that depended on the other directions? Unfortunately, this is also no good--in this setup, the curl of $E$ looks like: $$\nabla \times \vec{E} = \hat{x} \partial_y E_z - \hat{y} \partial_x E_z$$ Since we require $\nabla \times E = 0$ in electrostatics, both of these components must vanish and therefore there can be no spatial variation.

• This is not actually possible with an electrostatic field, due to the curllessness condition $\nabla\times\mathbf E=0$. (I'd be tempted to +1 for the chance to say curllessness in a sentence, though.) – Emilio Pisanty Apr 27 '15 at 18:13
• Wow, you're right. Somehow I was confident there was a problem that avoided this but it looks like not. Answer edited and wholly reversed. – zeldredge Apr 27 '15 at 18:20

If what you mean by a vacuum, something like outer space then electrostatic fields and electromagnetic waves do not need a medium to exist, (unlike sound waves and air pressure). In fact, the field from a proton from the solar wind in the vacuum of space would be constant in a reference frame moving with it, and there would be the inverse squared spatial variation.

• Hi Rabi, and welcome to the site! This does not actually address the specifics of the question (i.e. whether an electrostatic field in vacuum can have constant direction but variable magnitude). – Emilio Pisanty Apr 27 '15 at 18:15