Question

Now let's look at the potential gradient associated with a charged capacitor. Suppose we have a parallel-plate capacitor. The plates are separated by 6.0 mm

and carry charges of equal magnitude and opposite sign (Figure 1). The potential difference between the plates is 24.0 V
. Let the potential of the negatively charged plate be zero; then the potential of the positive plate is +
24.0 V
. Draw an enlarged sketch of (Figure 1); on it, show (1) the electric field lines between the plates and (2) the cross sections of the equipotential surfaces for which the potential is +
24.0 V
, +
18.0 V
, +
12.0 V
, +
6.0 V
, and 0.


For the capacitor in this example, find the magnitude of the electric field.
Express your answer with the appropriate units.
For the capacitor in this example, find the potential (again, assuming that V=0
at the negative plate) at a point 2.0 mm
from the negative plate and 4.0 mm
from the positive plate.

Answer 1

The student's query concerns calculations related to parallel-plate capacitors in high school physics, specifically capacitance, electric field, and the effect of dielectric materials on these properties.

Step-by-step explanation:

The question revolves around the concept of a parallel-plate capacitor, which is a common topic in high school physics. The student is asked to calculate various properties of capacitors under different conditions, such as potential difference, electric field, and capacitance changes when dielectrics are introduced or dimensions are altered.

Capacitance is a measure of a capacitor's ability to store charge per unit voltage, and the electric field is the force per unit charge in the region between the plates. Introducing a dielectric affects these properties by changing the ability of the capacitor to store charge. Equations relating to capacitance, electric field, and potential difference can be used to solve such problems for the different scenarios provided.