Shreyas Samudra said:
If one plate is at zero potential and the other one at V potential with respect to it, you need Vq work to move a charge q from the grounded plate to the non-grounded one.
But the potential is zero also at infinity, so the same work is needed to bring in the charge q from infinity to the non-grounded plate along its 'left side'.
You can not assume infinite plate. If you cut the universe into two, that plate still would be finite.
Assume finite plates instead, with lateral sizes much larger than the separation between them. In this case, you can assume finite charge densities, homogeneous far from the edges.
Assume two parallel plates of area A. Neither plates are grounded. The charge on the left plate is Q1 and the charge per unit area is σ1=Q1/A. Near the plates, the electric field due to the left plate would be -σ1/(2ε
0) on the left and σ1/(2ε
0) on the right (the blue arrows). The other plate has charge Q2, charge per unit area σ2=Q2/A, and electric field due to it is -σ2/(2ε
0) on the left and σ2/(2ε
0) on the right of it (the red arrows).
The resultant field is the sum of the contributions from both plates. On the left, it is E1=-(σ1+σ2)/(2ε
0). On the right, it is E2=(σ1+σ2)/(2ε
0).
Between the plates, the electric field is E=(σ1-σ2)/(2ε
0). The potential difference between the plates is V=(σ1-σ2)/(2ε
0)d.
Far away, the arrangement looks as a tiny charged object, of net charge Q1+Q2, and the electric field tends to zero as the distance increases.
Assume the right plate is grounded. Then E2=0, so the charge per unit area should be -σ1 on it. The electric field between the plates becomes E=σ1/ε
0, and the potential difference between the plates is V=σ1/ε
0d. You see, that the charge from the ground makes the net charge zero. Also E1=0.
You can extend this method to any number of parallel plates close to each other. The charge moving onto the grounded plate makes the whole arrangement neutral, the net charge of the whole arrangement becomes zero.