A parallel-plate capacitor is formed from two 1.0{\rm cm} \times 1.0{\rm cm} electrodes spaced 2.8{\rm mm} apart. The electric field strength inside the capacitor is 1.0\times10^{6}{N/C}}

part a What is the charge (in {\rm nC}) on positive electrode?
part b What is the charge (in {\rm nC}) on negative electrode?

C=εε₀A/d

ε=1
ε₀=8.85 •10⁻¹² F/m
A=1•1=1 cm²=1•10⁻⁴ m²
C=q/U=q/Ed
ε₀A/d= q/Ed
q=ε₀AE=8.85 •10⁻¹²•1•10⁻⁴•1•10⁶=
=8.85•10⁻¹⁰ C=0.885 nC
a. +0.885 nC
b. -0.885 nC

To find the charge on each electrode of the parallel-plate capacitor, we need to use the formula for the capacitance:

C = ε₀ * A / 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 each electrode, and d is the distance between the electrodes.

Given:
A = 1.0 cm x 1.0 cm = 1.0 x 10^-4 m²
d = 2.8 mm = 2.8 x 10^-3 m

First, let's calculate the capacitance:

C = (8.85 x 10^-12 F/m) * (1.0 x 10^-4 m²) / (2.8 x 10^-3 m)
C = 3.164 x 10^-12 F

Now, we can find the charge on each electrode using the formula:

Q = C * V

where Q is the charge, C is the capacitance, and V is the voltage.

Given that the electric field strength (E) inside the capacitor is 1.0 x 10^6 N/C, we can relate the voltage and electric field using:

E = V / d

Rearranging, we have:

V = E * d

V = (1.0 x 10^6 N/C) * (2.8 x 10^-3 m)
V = 2.8 x 10³ V

Now we can calculate the charges:

part a) Charge on the positive electrode:
Q = (3.164 x 10^-12 F) * (2.8 x 10³ V)
Q = 8.849 x 10^-9 C = 8.849 x 10^-3 nC

part b) Charge on the negative electrode:
According to the equation of conservation of charge, the sum of the charges on both electrodes should be zero. Thus, the charge on the negative electrode is equal in magnitude but opposite in sign to the charge on the positive electrode:

Q_negative = -Q_positive = -8.849 x 10^-3 nC

To find the charge on the positive electrode, we need to use the formula:

Q = CV

where Q is the charge, C is the capacitance, and V is the voltage. Since we are given the electric field strength (E) and the distance between the plates (d), we can use these values to find the voltage.

The electric field strength (E) is given as 1.0 × 10^6 N/C. This represents the voltage per unit distance, so we can write:

E = V/d

Rearranging the equation, we can solve for V:

V = E * d

Now we can substitute the given values:

V = (1.0 × 10^6 N/C) * (2.8 mm)

Converting the distance to meters:

V = (1.0 × 10^6 N/C) * (2.8 × 10^-3 m)

V = 2800 V

Now that we know the voltage, we can find the charge on the positive electrode by multiplying it by the capacitance. The capacitance (C) of a parallel-plate capacitor is given by:

C = ε₀ * (A/d)

where ε₀ is the permittivity of free space, A is the area of each plate, and d is the distance between the plates.

The area A is given as (1.0 cm) * (1.0 cm), but we need to convert it to square meters:

A = (1.0 × 10^-2 m) * (1.0 × 10^-2 m)

A = 1.0 × 10^-4 m²

Substituting the given values, we can find the capacitance:

C = (8.854 × 10^-12 F/m) * (1.0 × 10^-4 m²) / (2.8 × 10^-3 m)

C ≈ 3.169 × 10^-14 F

Finally, we can find the charge on the positive electrode:

Q = (3.169 × 10^-14 F) * (2800 V)

To convert the charge to nC (nanoCoulombs), we need to multiply by 10^9:

Q = (3.169 × 10^-5 C) * (10^9 nC/C)

Q ≈ 31.69 nC

Therefore, the charge on the positive electrode is approximately 31.69 nC.

To find the charge on the negative electrode, we can use the fact that in a neutral capacitor, the total charge is zero. Therefore:

Charge on the negative electrode = - Charge on the positive electrode

Thus, the charge on the negative electrode is also approximately 31.69 nC.