How does glass create an electric field?

I'm reading from Microchip's data sheet regarding their MAXTouch microchip, which details on how to create a touchscreen compatible with their chip, and in discussing the "mechanical stackup" of a touchscreen it is noted that

Front panel dielectric material has a direct bearing on sensitivity. Plastic front panels are usually suitable up to about 3 mm, and glass up to about 4 mm (dependent upon the screen size and layout). The thicker the front panel, the lower the signal-to-noise ratio of the measured capacitive changes and hence the lower the resolution of the touchscreen. In general, glass front panels are near optimal because they conduct electric fields almost twice as easily as plastic panels.

I'm unsure what this means that glass or plastic conducts electric fields. A typical mutual capacitive touchscreen consists of a set of x transmission lines underneath a set of y transmission lines (with space between them).

I thought that that space and the transmission lines is what causes the electric fields, not the glass which is on top of it all.

There's also a note which says

Care should be taken using ultra-thin glass panels as retransmission effects can occur, which can significantly degrade performance

So I guess this means that you wouldn't want to try and use a touch screen with no glass/plastic overlay at all?

• I expect that a material that "conducts electric fields" is an insulator. Conductors block electric fields so I guess plastic is about twice as conductive as glass (still not very). – immibis Jul 5 '18 at 1:14
• You don't want fingers in direct contact with the conductive material because that could conduct ESD to the circuitry. – Spehro Pefhany Jul 5 '18 at 12:19

This means is Glass ($\epsilon _R =4.7$) has twice the relative permittivity of plastic ( $\epsilon _R =2.3$ ) in the materials preferred here. This applies to both mutual and direct capacitance.
What is free space permittivity ? - about 9 picofarads per meter ( ~ 9 pF /m means means 9 pF would be the capacitance of two hands with a gap of $1/9^{th}$ of the hand width.