A parallel plate capacitor having plates 6.0 cm apart is connected across the terminals of a 12.0 V battery. (a) Being as quantitative as you can, describe the location and shape of the equipotential surface that is at a potential of +6.0 V relative to the potential of the negative plate. Avoid the edges of the plates. (b) DO the same for the equipotential surface that is at +2.0 V relative to the negative plate (c) what is the potential gradient between the plates?

ok..wouldn't the equipotential surface just be perpendicular to the electric field? or am I totally wrong?

If you call the negative terminal zero potential, the other terminal will be at +12. THus the six volt potential will be half the distance. The potential is a plane, ignoring edges. THe potential gradient? 12/.06 volts/meter.

The equitpotental surface is indeed perpendicular to the E field.

You are partially correct. The equipotential surfaces are perpendicular to the electric field lines, but in the case of a parallel plate capacitor, the electric field lines are uniform and perpendicular to the plates. Therefore, the equipotential surfaces will be parallel to the plates.

(a) For the surface at a potential of +6.0 V relative to the negative plate, the equipotential surface will be halfway between the two plates. It will be flat and parallel to the plates.

(b) For the surface at a potential of +2.0 V relative to the negative plate, the equipotential surface will be closer to the negative plate compared to the previous case. It will also be parallel to the plates but at a smaller distance from the negative plate.

(c) The potential gradient between the plates is the change in potential per unit distance. In this case, the potential difference between the plates is 12.0 V (positive plate to negative plate) and the distance between the plates is 6.0 cm (0.06 m). Therefore, the potential gradient can be calculated as:

Potential gradient = Potential difference / Distance
= 12.0 V / 0.06 m
= 200 V/m

So, the potential gradient between the plates is 200 volts per meter.

You are correct that the equipotential surfaces are perpendicular to the electric field. In the case of a parallel plate capacitor, the electric field is uniform and directed from the positive plate to the negative plate. So, each equipotential surface will be parallel to the plates.

To describe the location and shape of the equipotential surface that is at a potential of +6.0 V relative to the potential of the negative plate, we can start by considering the potential difference between the plates. In this case, the potential difference is 12.0 V (the voltage of the battery). The equipotential surfaces will be equally spaced out between the plates.

Since the potential difference between the plates is 12.0 V, and we are interested in the equipotential surface at +6.0 V relative to the negative plate, we can say that this equipotential surface is halfway between the plates.

The shape of the equipotential surface will be like that of a plane. It will be flat and parallel to the plates, running through the midpoint between them.

Similarly, for the equipotential surface that is at +2.0 V relative to the negative plate, it will be located one-third of the distance from the negative plate to the positive plate. This surface will also be flat and parallel to the plates.

Now, to determine the potential gradient between the plates, we can use the formula:

Electric field (E) = Potential difference (V) / Separation distance (d)

In this case, the separation distance between the plates is given as 6.0 cm. The potential difference between the plates is 12.0 V. Plugging these values into the formula, we can determine the electric field between the plates.

However, the question asks for the potential gradient, which is the rate of change of potential per unit distance. The potential gradient is equal to the magnitude of the electric field.

So, the potential gradient between the plates would be equal to the magnitude of the electric field.