On a whim I looked up whether magnets or electric fields can refract light, and while they generally can't, strong electric charges can:

Edit: there's probably no way to make an electric field strong enough on the macro scale to bend light, unless it passes by a black hole or magnetar, because the radius of the bending grows by 2nd power of charge but shrinks by the 4th power of distance from charge. But I'll leave my work here in case anyone is curious.

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See figure 1:

https://link.springer.com/article/10.1140/epjc/s10052-021-09...

  Equation 48:
  
  delta y = -E*(a^2)*(Q^2)
            -----------------
            80*pi*(m^4)*(b^4)
  
  E = 1 for parallel or (7/4)^2 for perpendicular?
  a = 137.036 (fine structure constant)
  m = 9.11e-31? (mass of electron? mass equivalent of electric field by E=mc^2?)
  Q = quantity of charge in coulombs
  b = smallest distance of light from point charge, or radius of light cone

Unfortunately the math is not written well IMHO, and it doesn't have any numeric examples, so the reader is forced to understand the entire paper before drawing conclusions.

It's conceivable that a strong charge placed millions of kilometers away could bend the laser light into a column again, although it might have to have an electric field close to the strength of an atom's, or 10^21 V/m. The breakdown voltage of space is 3x10^6 V/m, so it might require a black hole or high power to concentrate enough charge in one place, for example by using a ring of electron guns aimed at their center to simulate a focussed point charge.

But the bending is towards the charge and grows by Q^2, while falling by b^4. If m is the mass of the electron, then it's all multiplied by about 10^128, which suggests that a small charge would cause a large bend. Or if it's the mass equivalent, then a 1eV field might have an equivalent mass of (1.6x10-19 J)/(c^2) which is about 1/(10^36) or a multiplier of 10^144 ! But that doesn't sound right, so maybe someone can clarify it for us?

Edit: found another paper for calculating the bending angle of light in a nonuniform electric field (like near a point charge):

https://arxiv.org/abs/1012.1134

https://arxiv.org/pdf/1012.1134 (pdf)

Numeric example:

  As an example, for Z = 100, b = 10*lambda*e we get the bending angle theta = 3.4 × 10−8 radian for an x-ray of wavelength 5*lambda*e.

Probably a larger "impact parameter b, over which distance the bending occurs mostly" requires a proportionately larger electric field or point charge.

Edit: another paper calculating the bending of light in nonuniform electric fields near black holes:

https://arxiv.org/abs/1101.3433

https://arxiv.org/pdf/1101.3433 (pdf)

  Equation 18:
  
  delta y = -(E)(a^2)*(Q^2)*(lambda^4)
            --------------------------
            640*pi*e0*hbar*c*(b^4)
  
  E = 8 for parallel or 14 for perpendicular (substituted E for a to not conflict with alpha a)?
  a = -1 (doesn't say, but uses -1 in other examples)
  Q = quantity of charge in coulombs
  lambda = 2.426e−12 = 􏰀hbar/mc = the Compton length of the electron
  e0 = 9e9 = permitivity of free space
  hbar = 1.055e-34 = reduced Planck's constant
  c = 3e8 = speed of light
  b = smallest distance of light from point charge, or radius of light cone
  
  It grows by ((Q^2)*(lambda^4))/((e0*hbar*c)*(b^4))
  
  The top lambda^4 term works out to 10^-48 but the bottom e0*hbar\*c term works out to about 2.85e-16 so the formula only works for very small bend distance b.