Principle of mutual capacitive touch screen I'm curious about how the mutual capacitive touch screen works, so I did some research on internet and found an excellent explanation: youtube video.
According to the video, this is how the mutual capacitive touch screen works (if there is any misunderstanding, please let me know):


*

*The electric field btw transmit/receive plates leads to a polarization of human finger.

*As the finger tip approaches the plates, negative charges in finger tip repels the negative charges on Receive plate, but attracts more positive charges on Transmit plate.

*As a result, the mutual capacitance btw R and T plates decreases, and the sensor detect touch via sensing such drop of mutual capacitance.
My question is: why does an increase of $Q_T$ and decrease of $Q_R$ result in a drop of mutual capacitance?
 A: It's not so much the increase/decrease in charge that is responsible for a change in mutual conductance, as the change in geometry, since by definition the mutual capacitance of charged conductors is independent of their charge. This answer on the meaning of mutual capacitance provides details if needed. 
The interesting thing though is that the effect probably has to do also with other finer details of the design. To see why, consider a pair of overlapping receiving (R) and transmitting (T) plates, both held at fixed potentials $V_R$, $V_T$. In the absence of the user's finger the corresponding charges are related to the potentials and capacitances as 
$$
Q_R = C_R V_R + C_{RT}V_T
$$
$$
Q_T = C_{RT} V_R + C_TV_T
$$ 
where $C_R$, $C_T$ are plate self-capacitances and $C_{RT}$ is the mutual capacitance. When the user's finger (F) is brought in close proximity, and if the self-capacitances do not change, the plate charges vary according to 
$$
Q_R + \Delta Q_R = C_R V_R + \left(C_{RT}+\Delta C_{RT}\right) V_T + C_{RF}V_F
$$
$$
Q_T + \Delta Q_T = \left(C_{RT}+\Delta C_{RT}\right) V_R + C_TV_T + C_{TF}V_F
$$ 
while the finger charge should be $Q_F = C_{RF} V_R + C_{TF}V_T + C_FV_F$. Then the charge variations amount to
$$
\Delta Q_R = \Delta C_{RT} V_T + C_{RF}V_F\\
\Delta Q_T = \Delta C_{RT} V_R + C_{TF}V_F
$$
and give the change in mutual capacitance as 
$$
\Delta C_{RT} = \frac{\Delta Q_R - C_{RF}V_F}{V_T} = \frac{\Delta Q_T - C_{TF}V_F}{V_R} 
$$
If we use the last equality above to eliminate the finger's unknown potential $V_F$, we obtain 
$$
V_F = \frac{\Delta Q_R V_R - \Delta Q_T V_T}{C_{RF}V_R - C_{TF}V_T}
$$
and eventually
$$
\Delta C_{RT} = \frac{\Delta Q_R - C_{RF} \frac{\Delta Q_R V_R - \Delta Q_T V_T}{C_{RF}V_R - C_{TF}V_T}}{V_T} = \frac{C_{RF}\Delta Q_T - C_{TF}\Delta Q_R}{C_{RF}V_R - C_{TF}V_T}
$$
Note that up to now there is no assumption on the signs of the potentials and the charges. But assuming that $V_R$ and $V_T$ have well-defined polarity, say $V_R < 0$, $V_T >0$, the denominator in last expression has a well-defined sign too, in this case $C_{RF}V_R - C_{TF}V_T < 0$. This implies that if we are to have a decrease in $C_{RT}$, $\Delta C_{RT} < 0$,  then the denominator must have the opposite sign, respectively $C_{RF}\Delta Q_T - C_{TF}\Delta Q_R > 0$, giving
$$
\frac{\Delta Q_T}{C_{TF}} > \frac{\Delta Q_R}{C_{RF}}
$$
In other words, the user's finger must elicit (slightly) different responses in the receiving and the transmitting plates, and the design must provide such an effect accordingly. According to the info in the video you linked to, the finger's charge repels negative charges in the receiving plate, so $\Delta Q_R >0$, while attracting more positive charges in the transmitting plate, and $\Delta Q_T >0$ again. Since this makes both ratios above positive, the conclusion is that the design must provide for some subtle effect that makes the ratios different. 
Of course, this is a very simplified analysis based on the simplest possible model. The technical details are likely different. See for instance pg.22 of this "Touch Technologies Tutorial" for a much more sophisticated equivalent circuit.
