a large thin conducting plate has a net charge q.

the charge density on each side of conducting plate is : q/2A
electric field on each side is :q/2epsilon=q/4Aepsilon

Does the charge on the right surface contribute to the electric field to the left of the plate?

E on each side is indeed q/2epsiolon

No, E cannot transmit through a conductor.

No, the charge on the right surface of the conducting plate does not contribute to the electric field to the left of the plate.

To understand why, let's consider the concept of electric field and the behavior of a conducting plate.

The electric field is a vector field that describes the force experienced by a positive test charge placed in its vicinity. It is created by electric charges and can be influenced by the presence of other charges or conductors.

In the case of a conducting plate, the net charge q is distributed evenly on both sides, resulting in equal and opposite charge densities. The charge density on each side is q/2A, where A is the area of the plate.

When a conducting plate is in equilibrium, the electric field inside it is zero. This means that any excess charge present on the surface redistributes itself in such a way that cancels out the electric field inside the conductor. As a result, the charges redistribute themselves until the electric field inside the conductor becomes zero.

Now, if we consider the electric field outside the conductor, the presence of the conducting plate can have an effect. However, the electric field contributions from the charges on each side of the plate cancel each other out.

Since the electric field inside the conducting plate is zero, and the charges on either side are equal and opposite, the electric field created by the charges on the right surface will be canceled out by the electric field created by the charges on the left surface. Therefore, the charge on the right surface does not contribute to the electric field to the left of the plate.

In summary, while the presence of the conducting plate affects the electric field outside the plate, the charge on the right surface does not contribute to the electric field to the left of the plate.