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Travka [436]
3 years ago
13

Which of the following is a unit of volume of solids?

Physics
2 answers:
Verizon [17]3 years ago
6 0

Answer:

Cubic meters is the unit of volume.

Alla [95]3 years ago
4 0

Answer:

Cubic meters is the unit of volume.

Explanation:

Volume :

Volume is the multiplication of height, width and length of the solids.

In mathematically term,

V = w\times l\times d

Where, w = width

h = height

l = length

Put the unit into the formula

V = m\times m\times m

V =m^3

This is the unit of volume.

We know that,

Square meters is the unit of area.

Liters per cubic gram is the unit of density.

Gram per cubic centimeter is the unit of density in CGS.

Hence, Cubic meters is the unit of volume.

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There is a parallel plate capacitor. Both plates are 4x2 cm and are 10 cm apart. The top plate has surface charge density of 10C
liberstina [14]

Answer:

1) The total charge of the top plate is 0.008 C

b) The total charge of the bottom plate is -0.008 C

2) The electric field at the point exactly midway between the plates is 0

3) The electric field between plates is approximately 1.1294 × 10¹² N/C

4) The force on an electron in the middle of the two plates is approximately 1.807 × 10⁻⁷ N

Explanation:

The given parameters of the parallel plate capacitor are;

The dimensions of the plates = 4 × 2 cm

The distance between the plates = 10 cm

The surface charge density of the top plate, σ₁ = 10 C/m²

The surface charge density of the bottom plate, σ₂ = -10 C/m²

The surface area, A = 0.04 m × 0.02 m = 0.0008 m²

1) The total charge of the top plate, Q = σ₁ × A = 0.0008 m² × 10 C/m² = 0.008 C

b) The total charge of the bottom plate, Q = σ₂ × A = 0.0008 m² × -10 C/m² = -0.008 C

2) The electrical field at the point exactly midway between the plates is given as follows;

V_{tot} = V_{q1} + V_{q2}

V_q = \dfrac{k \cdot q}{r}

Therefore, we have;

The distance to the midpoint between the two plates = 10 cm/2 = 5 cm = 0.05 m

V_{tot} =  \dfrac{k \cdot q}{0.05} + \dfrac{k \cdot (-q)}{0.05}  = \dfrac{k \cdot q}{0.05} - \dfrac{k \cdot q}{0.05} = 0

The electric field at the point exactly midway between the plates, V_{tot} = 0

3) The electric field, 'E', between plates is given as follows;

E =\dfrac{\sigma }{\epsilon_0 } = \dfrac{10 \ C/m^2}{8.854 \times 10^{-12} \ C^2/(N\cdot m^2)} \approx 1.1294 \times 10^{12}\ N/C

E ≈ 1.1294 × 10¹² N/C

The electric field between plates, E ≈ 1.1294 × 10¹² N/C

4) The force on an electron in the middle of the two plates

The charge on an electron, e = -1.6 × 10⁻¹⁹ C

The force on an electron in the middle of the two plates, F_e = E × e

∴ F_e = 1.1294 × 10¹² N/C ×  -1.6 × 10⁻¹⁹ C ≈ 1.807 × 10⁻⁷ N

The force on an electron in the middle of the two plates, F_e ≈ 1.807 × 10⁻⁷ N

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