Question

A parallel-plate capacitor has capacitance C0 = 8.50 pF when there is air between the plates. The separation between the plates is 1.10 mm .

Part A
What is the maximum magnitude of charge that can be placed on each plate if the electric field in the region between the plates is not to exceed 3.00×104 V/m ?
Express your answer with the appropriate units.
Part B
A dielectric with K = 2.50 is inserted between the plates of the capacitor, completely filling the volume between the plates. Now what is the maximum magnitude of charge on each plate if the electric field between the plates is not to exceed 3.00×104 V/m ?
Express your answer with the appropriate units.

Answer 1

In a parallel-plate capacitor with air between the plates, the maximum magnitude of charge is determined by the capacitance and electric field. When a dielectric is inserted between the plates, the maximum magnitude of charge increases by a factor of the dielectric constant.

Step-by-step explanation:

Part A:

To calculate the maximum magnitude of charge that can be placed on each plate of a parallel plate capacitor, we need to use the formula for capacitance:

C = ε₀A/d

where C is the capacitance, ε₀ is the permittivity of free space (8.85 x 10^-12 F/m), A is the area of the plates, and d is the separation between them.

Given that the capacitance C₀ is 8.50 pF (8.50 x 10^-12 F) and the separation d is 1.10 mm (1.10 x 10^-3 m), we can rearrange the formula to solve for the maximum magnitude of charge Q:

Q = C₀V

where V is the electric field.

Substituting the values, we get:

Q = (8.50 x 10^-12 F)(3.00 x 10^4 V/m)

Part B:

When a dielectric with a dielectric constant K is inserted between the plates, the capacitance increases by a factor of K.

So, the new maximum magnitude of charge Q' is given by:

Q' = KQ

Substituting K = 2.50, we can determine the new maximum magnitude of charge Q'.

Answer 2

The maximum magnitude of charge that can be placed on each plate of a parallel-plate capacitor can be found using the formula Q = CV. When there is air between the plates, the maximum magnitude of charge is 2.80 × 10-10 C. When a dielectric with a dielectric constant K is inserted between the plates, the maximum magnitude of charge becomes 7.00 × 10-10 C.

Step-by-step explanation:

Part A:

To find the maximum magnitude of charge that can be placed on each plate of a parallel-plate capacitor, we can use the formula Q = CV, where Q is the charge, C is the capacitance, and V is the potential difference.

Given:

• C0 = 8.50 pF = 8.50 × 10-12 F (capacitance when there is air between the plates)
• E = 3.00 × 104 V/m (maximum electric field strength)
• d = 1.10 mm = 1.10 × 10-3 m (separation between the plates)

Using the formula for electric field, E = V/d, we can rearrange to find the potential difference V = Ed.

V = (3.00 × 104 V/m) × (1.10 × 10-3 m) = 33.00 V

Now we can use the formula Q = CV to find the charge:

Q = (8.50 × 10-12 F) × (33.00 V) = 2.80 × 10-10 C

Therefore, the maximum magnitude of charge that can be placed on each plate is 2.80 × 10-10 C.

Part B:

When a dielectric with a dielectric constant K is inserted between the plates of the capacitor, the capacitance changes according to the formula C = KC0. The potential difference between the plates remains the same, so we can use the formula Q = CV to find the charge with the new capacitance.

Given:

• K = 2.50 (dielectric constant)
• E = 3.00 × 104 V/m (maximum electric field strength)
• d = 1.10 mm = 1.10 × 10-3 m (separation between the plates)

The capacitance with the dielectric is C = KC0 = (2.50) × (8.50 × 10-12 F) = 2.12 × 10-11 F.

Using the formula Q = CV, we can find the charge:

Q = (2.12 × 10-11 F) × (33.00 V) = 7.00 × 10-10 C

Therefore, the maximum magnitude of charge that can be placed on each plate with the dielectric is 7.00 × 10-10 C.