What is a personal bias?

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

4 0

2 years ago

Metals in group 2 on the period table most commonly form which type

LuckyWell [14K]

Answer:

alkaline earth metals

Group 2 metals, the alkaline earth metals, have 2 valence electrons, and thus form M2+ ions. The halogens, Group 17 , reach a full valence shell upon reduction, and thus form X− ions

Explanation:

6 0

3 years ago

A person holding a lunch bag is moving upward in a hot air balloon at a constant speed of 7.3 m/s . When the balloon is 24 m abo

Kitty [74]

Explanation:

Given that,

Initial speed of the bag, u = 7.3 m/s

Height above ground, s = 24 m

We need to find the speed of the bag just before it reaches the ground. It can be calculated using third equation of motion as :

$v^2=u^2+2as$

$v^2=(7.3)^2+2\times 9.8\times 24$

$v=\sqrt{523.69}$

v = 22.88 m/s

So, the speed of the bag just before it reaches the ground is 22.38 m/s. Hence, this is the required solution.

8 0

2 years ago

You are traveling 70 mph (31.3 m/s) and slam on the brakes to avoid hitting another car. How far do you travel if it takes you 8

Arlecino [84]

You traveled a distance of 620.075 meters if it takes you 8.5 seconds to stop.

Given the following data:

Initial velocity, U = 31.3 m/s

Time, t = 8.5 seconds.

We know that acceleration due to gravity (a) for an object is equal to 9.8 meter per seconds square.

To find the distance traveled, we would use the second equation of motion:

Mathematically, the second equation of motion is given by the formula;

$S = ut + \frac{1}{2} at^2$

Where:

S is the distance travelled.

u is the initial velocity.

a is the acceleration.

t is the time measured in seconds.

Substituting the parameters into the formula, we have;

$S = 31.3(8.5) + \frac{x}{y} (9.8)(8.5^2)\\\\S = 266.05 + 4.9(72.25)\\\\S = 266.05 + 354.025$

Distance, S = 620.075 meters.

Therefore, you traveled a distance of 620.075 meters if it takes you 8.5 seconds to stop.