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3 years ago

12

The potential difference between the plates of a parallel plate capacitor is 4.00 V. If the plate separation for this capacitor

is 6.00 cm, what is the intensity of the electric field between the plates of this capacitor?

1 answer:

shtirl [24]3 years ago

4 0

<h2>Answer:</h2>

66.67 V / m

<h2>Explanation:</h2>

The electric field intensity (E) between the plates of a capacitor is the ratio of the potential difference (V) between the plates to the distance (d) of separation between the plates. Electric field intensity can be measured in V/m. It can be written as follows;

E = V / d ----------------------------(i)

<em>From the question;</em>

V = 4.00V

d = 6.00cm = 0.06m

<em>Substitute these values into the equation above. i.e equation (i) as follows;</em>

E = 4.00 / 0.06

E = 66.67 V / m

Therefore, the electric field intensity between the plates of the capacitor is 66.67 V / m

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Two particles move along an x axis. The position of particle 1 is given by x ! 6.00t 2 # 3.00t # 2.00 (in meters and seconds); t

s344n2d4d5 [400]

Answer:

15.6m/s

Explanation:

V1= $\frac{dx}{dt}=\frac{d}{dt}(6t^{2}+3t+2)$ because the derivate of the position is the velocity

V1=12t+3

V2=20+ $\int\limits^_ {} \,$ -8t because the integral of the acceleration is the velocity

V2= $20-4t^{2}$

V1=V2 to see when the velocities of particles match

12t+3=20-4t^2

4t^2+12t-17=0 we resolve this with $\frac{-b+-(\sqrt{b^{2} -4ac} )}{2a}$

and we take the positif root

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7 0

3 years ago

natulia [17]

Answer:

$\% Error = 2.6\%$

Explanation:

Given

$x: 1.54, 1.53, 1.44, 1.54, 1.56, 1.45$

Required

Determine the percentage error

First, we calculate the mean

$\bar x = \frac{\sum x}{n}$

This gives:

$\bar x = \frac{1.54+ 1.53+ 1.44+ 1.54+ 1.56+ 1.45}{6}$

$\bar x = \frac{9.06}{6}$

$\bar x = 1.51$

Next, calculate the mean absolute error (E)

$|E| = \sqrt{\frac{1}{6}\sum(x - \bar x)^2}$

This gives:

$|E| = \sqrt{\frac{1}{6}*[(1.54 - 1.51)^2 +(1.53- 1.51)^2 +.... +(1.45- 1.51)^2]}$

$|E| = \sqrt{\frac{1}{6}*0.0132}$

$|E| = \sqrt{0.0022}$

$|E| = 0.04$

Next, calculate the relative error (R)

$R = \frac{|E|}{\bar x}$

$R = \frac{0.04}{1.51}$

$R = 0.026$

Lastly, the percentage error is calculated as:

$\% Error = R * 100\%$

$\% Error = 0.026 * 100\%$

$\% Error = 2.6\%$

4 0

3 years ago

An student writes a bird that is divining for prey has speed of _ 10m/s what is wrong with the statement.

Anika [276]

Answer:

The description including its issue is summarized throughout the clarification portion below.

Explanation:

Speed would be a measurement of size. Which has only significance and therefore no path or guidance. Student says acceleration of -10m/s, therefore this assumption is incorrect because this same negative correlation indicates the path.
In reality, this same participant defined the speed (which seems to have two very different motion of an object).

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3 years ago

Two teams are playing tug-of-war. The team on the right is pulling with 4320 N. The team on the left is pulling with 4380 N. Whi

zmey [24]

60 N to the left because 4380-4320is 60

6 0

3 years ago

Two slits in an opaque barrier each have a width of 0.020 mm and are separated by 0.050 mm. When coherent monochromatic light pa

Semenov [28]

Answer:

The answer is 5

Explanation:

The maximum interference is:

m * λ = d * sinθi

Where m = 0,1,2,3,...

The first minimum diffraction is:

λ = a * sinθd

|sinθi| < sinθd

Where

(|m| * λ)/d < λ/a

|m| < d/a = 2.5

|m|max = 2

It can be concluded that coherent monochromatic light passes through the slits, therefore the maximum number of interference is 5.

5 0

4 years ago