Two circular plates, each with a radius of 8.22 cm, have equal and opposite charges of magnitude 3.052 μc. Calculate the electric field between the two plates. assume that the separation distance is small in comparison to the diameter of the plates. electric field: n/c the plates are slowly pulled apart, doubling the separation distance. again, assume the separation distance remains small in comparison to the diameter of the plates. what changes occur with the electric field between the plates? the electric field decreases by a factor of 2. the electric field stays the same. the electric field increases by a factor of 2. the electric field decreases by a factor of 4. the electric field increases by a factor of 4.
The Correct Answer and Explanation is:
To determine the electric field between two parallel plates with equal and opposite charges, we can use the concept of a uniform electric field created by charged plates in close proximity. For this situation, we assume the separation between the plates is much smaller than the diameter of the plates, allowing us to treat the electric field as uniform.
Step 1: Calculating the Initial Electric Field
The electric field ( E ) between two parallel plates with surface charge densities ( \sigma ) (charge per unit area) is given by:
[
E = \frac{\sigma}{\epsilon_0}
]
where:
- ( \sigma = \frac{Q}{A} ),
- ( Q ) is the charge magnitude on each plate,
- ( A ) is the area of each plate, and
- ( \epsilon_0 ) is the permittivity of free space (( \approx 8.85 \times 10^{-12} \, \text{C}^2/\text{N} \cdot \text{m}^2 )).
Since each plate is circular with a radius ( r = 8.22 \, \text{cm} = 0.0822 \, \text{m} ):
[
A = \pi r^2 = \pi (0.0822 \, \text{m})^2 \approx 0.0212 \, \text{m}^2
]
The surface charge density ( \sigma ) is:
[
\sigma = \frac{Q}{A} = \frac{3.052 \times 10^{-6} \, \text{C}}{0.0212 \, \text{m}^2} \approx 1.44 \times 10^{-4} \, \text{C/m}^2
]
Substituting ( \sigma ) and ( \epsilon_0 ) into the electric field equation:
[
E = \frac{1.44 \times 10^{-4} \, \text{C/m}^2}{8.85 \times 10^{-12} \, \text{C}^2/\text{N} \cdot \text{m}^2} \approx 1.63 \times 10^7 \, \text{N/C}
]
Step 2: Effect of Doubling the Separation
When the plates are slowly pulled apart, doubling the separation, the electric field between parallel plates with equal and opposite charges remains the same, provided the separation distance is still small relative to the diameter of the plates. This is because the electric field ( E = \frac{\sigma}{\epsilon_0} ) depends only on the surface charge density ( \sigma ) and ( \epsilon_0 ), both of which remain constant.
Thus, the correct answer is that the electric field stays the same. Unlike the electric field generated by a point charge, which weakens with increased distance, the electric field between two parallel plates remains uniform and unaffected by their separation (as long as the separation is much smaller than the plate diameter). This is a characteristic of the ideal parallel plate capacitor setup and makes it unique compared to fields generated by point charges.