A large power plant generates electricity at 12.0 kV. Its old transformer once converted the voltage to 390 kV. The secondary of this transformer is being replaced so that its output can be 515 kV for more efficient cross-country transmission on upgraded transmission lines. (a) What is the ratio of turns in the new secondary compared with turns in the old secondary

Answer:

neither will happen

Explanation:

cause the water is already defreezed

Answer:

8.9*10^6 V/m

Explanation:

The expression for electric field strength E is given as

[tex]E=\frac{V}{d}[/tex]

where V= voltage

           d= distance of separation

Given data

[tex]voltage= 84 mV= 84*10^-^3\\distance= 9.4*10^-^9[/tex]

substituting our given data into the electric field strength formula we have

[tex]E= \frac{84*10^-^3}{ 9.4*10^-^9} \\\\E= \frac{84}{9.4} *10^-^3^-^(^-^9^)\\\\E= \frac{84}{9.4} *10^-^3^+^9\\\\E= \frac{84}{9.4} *10^6\\\\E=8.9*10^6 V/m[/tex]

Answer:

b)using

DT = (LAl - LBr) / (LBr aBr - LAl aAl)

DT = (10.02-10)/(10*19x10-6 –10.02*24x10-6)

DT = -396 C°

20°C + -396 C° < -273.15 °C;

So the temperature will be -396° this is unattainable because we can’t go below absolute zero

Answer:

   h = 4 sin (314.15 t)

Explanation:

This is a kinematics exercise, as the system is rotating at a constant speed.

          w = θ / t

          θ = w t

in angular motion all angles are measured in radians, which is defined

         θ = s / R

   we substitute

          s / R = w t

          s = w R t

let's reduce the magnitude to the SI system

    w = 3000 rev / min (2π rad / 1rev) (1min / 60 s) = 314.16 rad / s

let's calculate

   s = 314.16 4 t

   s = 1,256.6 t

this is the value of the arc

Let's find the function of this system, let's use trigonometry to find the projection on the x axis

                  cos θ = x / R

                  x = R cos θ

                  x = R cos wt

projection onto the y-axis is

               sin θ = y / R

how is a uniform movement

               θ = w t

               y = R sin wt

In the case y = h

              h = R sin wt

              h = 4 sin (314.15 t)

Answer:

Explanation:

E= −L ΔI / Δt.

L = E Δt / ΔI

Hence the unit of inductance may be V s A⁻¹

or volt s per ampere .

In the given case

change in current ΔI = - 2.5 A

change in time = .015 s

L = .56 H

E = − L ΔI / Δt.

= .56 x 2.5 / .015

= 93.33 V .

Answer:

(a) 3.9cm

(b) 1.66 x 10⁻⁸s

Explanation:

Since the electron is moving in a circular path, the centripetal acceleration needed to keep it from slipping off is provided by the magnetic force. This force (F), according to Newton's second law of motion is given by,

F = m x a          --------------(i)

Where;

m = mass of the particle

a = acceleration of the mass

The centripetal acceleration is given by;

a = v² / r          [v = linear velocity of particle, r = radius of circular path]

Therefore, equation (i) becomes;

F = m v²/ r             --------------------(ii)

The magnitude of the magnetic force on a moving charge in a magnetic field as stated by Lorentz's law is given by;

F = qvBsinθ          -------------(iii)

Where;

q = charge of the particle

v = velocity of the particle

B = magnetic field

θ = angle between the velocity and the magnetic field

Combine equations (ii) and (iii) as follows;

m (v² / r) = qvBsinθ         [divide both side by v]

m v / r = qBsinθ              [make r subject of the formula]

r = (m v) / (qBsinθ)              ---------(iv)

(a) From the question;

v = 1.48 x 10⁷m/s

B = 2.14mT = 2.14 x 10⁻³T

θ = 90°          [since the direction of velocity is perpendicular to magnetic field]

m = mass of electron = 9.11 x 10⁻³¹kg

q = charge of electron = 1.6 x 10⁻¹⁹C

Substitute these values into equation (iv) as follows;

r = (9.11 x 10⁻³¹ x 1.48 x 10⁷) / (1.6 x 10⁻¹⁹ x 2.14 x 10⁻³ sin 90°)

r = 3.9 x 10⁻²m

r = 3.9cm

Therefore, the radius of the circular path is 3.9cm

(b) The time interval required to complete one revolution is the period (T) of the motion of the electron and it is given by

T = d / v          --------------(*)

Where;

d = distance traveled in the circular path in one complete turn = 2πr

v = velocity of the motion = 1.48 x 10⁷m/s

d = 2 π (3.9 x 10⁻²)            [Take π = 22/7 = 3.142]

d = 2(3.142)(3.9 x 10⁻²) = 0.245m

Substitute the values of d and v into equation (*) as follows;

T = 0.245 / 1.48 x 10⁷

T = 0.166 x 10⁻⁷s

T = 1.66 x 10⁻⁸s

Therefore, the time interval is 1.66 x 10⁻⁸s

Answer:

The frequency increases by 4 because it is inversely proportional to the wavelength.

Answer:

Maximum height, h = 3062.5m

Total time of flight, T = 49.49secs or 50secs approx.

Range, R = 12250m

Explanation:

Given data:

U = 350m/s

Angle = 45°

Assume g = 10m/s

At the greatest height, v = 0

Therefore,

V^2 = U^2 sin^2 × angle - 2×g×h

Substituting values:

0^2 = 350^2 sin^2 (45) - 2 × 10 × h

Let h = maximum height reached

Rearranging gives:

350^2 sin^2(45) = 2 x 10 x h

h = 350^2 sin^2(45)/2×10

h = 122500 x 0.5/20

h = 61250/20

h = 3062.5m

2)Total time of flight, T

T = 2U sin(angle)/g

= 2x350 sin(45)/10

= 494.9747/10

= 49.49secs or 50sec approx.

3) Range of projectile, R

R = U^2 sin2(angle)

= 350^2 sin2 (45)

= 122500 x 1/10

= 12250m

Answer:

The induced voltage is  [tex]\epsilon = 4.53 \ V[/tex]

Explanation:

From the question we are told that

    The  number of turns is  [tex]N = 1300 \ turns[/tex]

     The diameter is  [tex]d = 2.2 \ cm =0.022 \ m[/tex]

     The  initial magnetic field is  [tex]B_i = 0.11 \ T[/tex]

     The final magnetic field is  [tex]B_f = 0 \ T[/tex]

    The time taken is  [tex]t = 12 \ ms = 12*10^{-3} \ s[/tex]

The radius is mathematically evaluated as

         [tex]r = \frac{d}{2}[/tex]

substituting values

         [tex]r = \frac{0.022}{2}[/tex]

         [tex]r = 0.011 \ m[/tex]

Generally the induced emf  is mathematically represented as

      [tex]\epsilon = - N * \frac{d\phi}{dt}[/tex]

Where  [tex]d\phi[/tex] is the change in magnetic flux of the wire which is mathematically represented as

      [tex]d \phi = dB* A * cos \theta[/tex]

=>  [tex]d \phi = (B_f - B_i )* A * cos \theta[/tex]

Here  [tex]\theta = 0[/tex]

since the axis of the coil is parallel to the field

    Where A  is the cross-sectional area of the coil which is mathematically represented as

      [tex]A = \pi * r^2[/tex]

       [tex]A = 3.142 * 0.011^2[/tex]

      [tex]A = 3.80*10^{-4} \ m^2[/tex]

So the induced emf

        [tex]\epsilon = - 1300 * \frac{(0- 0.11) * 3.80*10^{-4}}{12*10^{-3}}[/tex]   Here we substituted the values of  [tex]d \phi[/tex]

       [tex]\epsilon = 4.53 \ V[/tex]

The emf induced in the coil at the given magnetic field strength is 4.53 V.

The given parameters;

The area of the coil is calculated as follows;

[tex]A = \pi r^2 = \frac{\pi d^2}{4} \\\\A = \frac{\pi \times 0.022^2}{4} = 0.00038 \ m^2[/tex]

The emf induced in the coil is calculated as follows;

[tex]emf = -N\frac{d\phi}{dt} \\\\emf = N (\frac{\phi_1 - \phi_2}{t} )\\\\emf = N(\frac{AB_1 - AB_2}{t} )\\\\emf = NA(\frac{B_1 - B_2}{t} )\\\\emf = 1300 \times 0.00038 (\frac{0.11 - 0}{12 \times 10^{-3}} )\\\\emf = 4.53 \ V[/tex]

Thus, the emf induced in the coil at the given magnetic field strength is 4.53 V.

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Answer:

A- to simplify the study of things

Explanation:

a visual reference

Answer:

Explanation:

The sum of force along both components x and y is expressed as;

[tex]\sum Fx = ma_x \ and \ \sum Fy = ma_y[/tex]

The magnitude of the net force which is also known as the resultant will be expressed as [tex]R =\sqrt{(\sum Fx)^2 + (\sum Fx )^2}[/tex]

To get the resultant, we need to get the sum of the forces along each components. But first lets get the acceleration along the components first.

Given the position of the object along the x-component to be x = 6t² − 4;

[tex]a_x = \frac{d^2 x }{dt^2}[/tex]

[tex]a_x = \frac{d}{dt}(\frac{dx}{dt} )\\ \\a_x = \frac{d}{dt}(6t^{2}-4 )\\\\a_x = \frac{d}{dt}(12t )\\\\a_x = 12m/s^{2}[/tex]

Similarly,

[tex]a_y = \frac{d}{dt}(\frac{dy}{dt} )\\ \\a_y = \frac{d}{dt}(5t^{3} +6 )\\\\a_y = \frac{d}{dt}(15t^{2} )\\\\a_y = 30t\\a_y \ at \ t= 2.15s; a_y = 30(2.15)\\a_y = 64.5m/s^2[/tex]

[tex]\sum F_x = 3.15 * 12 = 37.8N\\\sum F_y = 3.15 * 64.5 = 203.18N[/tex]

[tex]R = \sqrt{37.8^2+203.18^2}\\ \\R = \sqrt{1428.84+41,282.11}\\ \\R = \sqrt{42.710.95}\\ \\R = 206.67N[/tex]

Hence, the magnitude of the net force acting on this object at t = 2.15 s is approximately 206.67N

Answer:

Radiation is transferred through electromagnetic waves so D.

Explanation:

Answer:

D. Waves

Explanation:

a and b don't make much sense, conduction is transfer of energy through touch

Answer:

(a) 4.334 × 10¹¹ joules are required to propel the rocket an unlimited distance away from Earth's surface, (b) The rocket has travelled 3999.865 miles from the Earth's surface with the half of the total work.

Explanation:

The complete statement is: "Use the weight of the rocket to answer the question. (Use 4000 miles as the radius of Earth and do not consider the effect of air resistance.) 7 metric ton rocket (a) How much work is required to propel the rocket an unlimited distance away from Earth's surface, (b) How far has the rocket traveled when half the total work has occurred?"

(a) The work required to propel the rocket is given by the change in gravitational potential energy, whose expression derives is described below:

[tex]U_{g, f} - U_{g, o} = -G\cdot M\cdot m \cdot \left[\frac{1}{r_{f}}-\frac{1}{r_{o}} \right][/tex]

Where:

[tex]U_{g,o}[/tex], [tex]U_{g,f}[/tex] - Initial and final gravitational potential energies, measured in joules.

[tex]m[/tex], [tex]M[/tex] - Masses of the rocket and planet Earth, measured in kilograms.

[tex]G[/tex] - Universal gravitation constant, measured in newton-square meters per square kilogram.

[tex]r_{o}[/tex], [tex]r_{f}[/tex] - Initial and final distances of the rocket with respect to the center of the Earth, measured in meters.

The initial distance and rocket mass are converted to meters and kilograms, respectively:

[tex]r_{o} = (4000\,mi)\cdot \left(1609.34\,\frac{m}{mi} \right)[/tex]

[tex]r_{o} = 6,437,360\,m[/tex]

[tex]m = (7\,ton)\cdot \left(1000\,\frac{kg}{ton} \right)[/tex]

[tex]m = 7000\,kg[/tex]

Given that [tex]m = 7000\,kg[/tex], [tex]M = 5.972\times 10^{24}\,kg[/tex], [tex]G = 6.674\times 10^{-11}\,\frac{N\cdot m^{2}}{kg^{2}}[/tex], [tex]r_{o} = 6,437,360\,m[/tex] and [tex]r_{f} \rightarrow +\infty[/tex], the work equation is reduced to this form:

[tex]U_{g,f} - U_{g,o} = \frac{G\cdot m \cdot M}{r_{o}}[/tex]

[tex]U_{g,f} - U_{g,o} = \frac{\left(6.674\times 10^{-11}\,\frac{N\cdot m^{2}}{kg^{2}} \right)\cdot (7000\,kg)\cdot (5.972\times 10^{24}\,kg)}{6,437,360\,m}[/tex]

[tex]U_{g,f} - U_{g,o} = 4.334\times 10^{11}\,J[/tex]

4.334 × 10¹¹ joules are required to propel the rocket an unlimited distance away from Earth's surface.

(b) The needed change in gravitational potential energy is:

[tex]U_{g,f} - U_{g,o} = 2.167\times 10^{11}\,J[/tex]

The expression for the change in gravitational potential energy is now modified by clearing the final distance with respect to the center of Earth:

[tex]U_{g, f} - U_{g, o} = -G\cdot M\cdot m \cdot \left[\frac{1}{r_{f}}-\frac{1}{r_{o}} \right][/tex]

[tex]\frac{U_{g,o}-U_{g,f}}{G\cdot M \cdot m} = \frac{1}{r_{f}} - \frac{1}{r_{o}}[/tex]

[tex]\frac{1}{r_{f}} = \frac{1}{r_{o}} + \frac{U_{g,o}-U_{g,f}}{G\cdot M\cdot m}[/tex]

[tex]r_{f} = \left(\frac{1}{r_{o}} + \frac{U_{g,o}-U_{g,f}}{G\cdot M\cdot m} \right)^{-1}[/tex]

If [tex]m = 7000\,kg[/tex], [tex]M = 5.972\times 10^{24}\,kg[/tex], [tex]G = 6.674\times 10^{-11}\,\frac{N\cdot m^{2}}{kg^{2}}[/tex], [tex]r_{o} = 6,437,360\,m[/tex]  and [tex]U_{g,f} - U_{g,o} = 2.167\times 10^{11}\,J[/tex], then:

[tex]r_{f} = \left[\frac{1}{6,437,360\,m}-\frac{2.167\times 10^{11}\,J}{\left(6.674\times 10^{-11}\,\frac{N\cdot m^{2}}{kg^{2}} \right)\cdot (7000\,kg)\cdot (5.972\times 10^{24}\,kg)} \right]^{-1}[/tex]

[tex]r_{f} \approx 12,874,502.49\,m[/tex]

The final distance with respect to the center of the Earth in miles is:

[tex]r_{f} = (12,874,502.49\,m)\cdot \left(\frac{1}{1609.34}\,\frac{mi}{m} \right)[/tex]

[tex]r_{f} = 7999.865\,mi[/tex]

The distance travelled by the rocket is: ([tex]r_{f} = 7999.865\,mi[/tex], [tex]r_{o} = 4000\,mi[/tex])

[tex]\Delta r = r_{f}-r_{o}[/tex]

[tex]\Delta r = 7999.865\,mi - 4000\,mi[/tex]

[tex]\Delta r = 3999.865\,mi[/tex]

The rocket has travelled 3999.865 miles from the Earth's surface with the half of the total work.

Answer:

64 times

Explanation:

if increase of 1 gives you 32

then increase of 2 will give you its double

64

If you increase one, you get 32 then multiplying by 2 will give you 64, which is its double.

An earthquake is a sudden energy released in the Earth's lithosphere that causes shock wave, which cause the Earth's surface to shake. Earthquakes can range in strength from ones that are so small that no one can feel them to quakes that are so powerful that they uproot entire cities, launch individuals and objects into the air, and harm vital infrastructure.

The frequency, kind, and intensity of earthquakes observed over a specific time period are considered to be the seismic activity of an area.

The average rate of earthquake energy output per unit volume determines the basicity of a certain area of the Earth. The non-earthquake seismic rumbling is also alluded to as a tremor.

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Answer:

When we have completely polarized light with intensity I0, and it passes through a polarizing filter with its axis at an angle θ with respect to the light's polarization direction, the new intensity of the light will be:

I = I0*cos^2(θ)

This is called the "Malus' law".

in this case, we have:

I0 =  125 W/m^2

θ = 89.5°

then:

I = (125 W/m^2)*cos^2(89.9°) = 0.00038 W/m^2

Answer:

4°C

Explanation:

Water is densest at 4°C.  Since dense water sinks, the bottom of the lake will be 4°C.

Answer:

W = 1014 J = 1.014 KJ

Explanation:

As, Sam has to stop the boats in the log ride. Therefore, the work Sam needs to do, in order to stop a boat must be equal to the kinetic energy of the boat:

Work Done by Sam = Kinetic Energy of the Boat

W = K.E

W = (1/2)mv²

where,

m = mass of boat and its rider = 1200 kg

v = speed of the boat = 1.3 m/s

Therefore,

W = (1/2)(1200 kg)(1.3 m/s)²

W = 1014 J = 1.014 KJ

Answer:

The bright fringes will appear much closer together

Explanation:

Because λn = λ/n ,

And the wavelength of light in water is smaller than the wavelength of light in air. Given that the distance between bright fringes is proportional to the wavelength

Answer:

10 kgm/s

Explanation:

Change in momentum: This can be defined as the product of mass and change in velocity. The S.I unit of change in momentum is Kgm/s.

From the question,

ΔM = m(v-u)...................... Equation 1

Where ΔM = change in momentum, u = initial velocity, v = final velocity.

Note: Let upward direction be negative, and downward direction be positive.

Given: m = 0.2 kg, v = -20 m/s, u = 30 m/s

Substitute into equation 1

ΔM = 0.2(-20-30)

ΔM = 0.2(-50)

ΔM = -10 kgm/s.

The negative sign shows that the change in momentum is Upward

The magnitude of the change in momentum of the ball as a result of the collision with the sidewalk is -10 kg-m/s.

Given data:

The mass of rubber ball is, m = 0.2 kg.

The initial speed of ball is, u = 30 m/s.

The final rebounding speed of ball is, v = - 20 m/s ( Negative sign shows that during the rebounding, the ball changes its direction)

The momentum of any object is defined as the product of mass and change in velocity. The S.I unit of momentum is Kg-m/s. And the expression for the change in momentum is given as,

[tex]p= m ( v-u)[/tex]

Solving as,

[tex]p= 0.2 \times ( -20-30)\\\\p=-10 \;\rm kg.m/s[/tex]

Thus, we can conclude that the magnitude of the change in momentum of the ball as a result of the collision with the sidewalk is -10 kg-m/s.

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Answer:

1) c. Helium

2) Iron

3) False.

Explanation:

1. Red dwarf is the smallest and the coolest star on the sequence. These are common stars in the milky way. Red dwarfs contains metals and the elements with higher atomic number. It is found that Helium is produced in red dwarf stars.

2. Iron is the highest atomic number element that is produced in cores of largest stars. The highest mass stars can make all elements up to iron, which is the heaviest element they can produce.

3. The end of stars life is dependent on the mass they are born with. It is not necessary that all red dwarf stars will become white dwarf stars faster than sun like star.

Answer:

0.11m

Explanation:

let's assume the boat is of uniform construction

Ignoring friction losses

Also assume the origin is at the end of the boat originally with the heavier person

the center of mass of the whole system will not change relative to the water when the two swap ends

Originally, the center of mass is

85[0] + 90[3.5/2] + 50[3.5] / (85 + 90 + 50) = 1.14m from the origin

after the swap, the center of mass is

50[0] + 90[3.5/2] + 85[3.5] / (85 + 90+ 50) = 1.030m from the origin

The center of mass has shifted

1.14-1.030 = 0.11m

as no external force acted on the system, the center of mass relative to the water will not change. The boat will therefore shift towards the end where the heavier person originally sat

Answer:

Final velocity (v) = 0.509 m/s (Approx)

Explanation:

Ant use impulse power

Given:

Mass of ant = 12 mg = 12 × 10⁻⁶ kg

Average force = 47 mN = 47 × 10⁻³ N

Initial velocity(u) = 0

Time taken = 0.13 ms = 0.13 × 10⁻³ s

Find:

Final velocity (v)

Computation:

Force × Time =  change in momentum

(47 × 10⁻³ N)(0.13 × 10⁻³ s) = mv - mu

(47 × 10⁻³ N)(0.13 × 10⁻³ s) = m(v - u)

6.11 × 10⁻⁶ = 12 × 10⁻⁶(v - 0)

6.11 = 12 v

Final velocity (v) = 0.509 m/s (Approx)

Answer:

dsinФ=mΔ

d=1/N

d=1/3500*[tex]10^{-2} m\\[/tex]

d=2.8*[tex]10^{-6}[/tex]

NOw apply all values on formula

dsinФ=mΔ

2.8*[tex]10^{-6\\}[/tex]sinФ=5*589*[tex]10^{-9}[/tex]

sinФ=1.05 error

so due fifth maximum order it cannot be soved by this grating

Answer:

E = 12.25 x 10³ N/C = 12.25 KN/C

Explanation:

In order to balance the weight of the object the electrostatic force due to the electric field must be equal to the weight of the body or charge. Therefore,

Electrostatic Force = Weight

E q = mg

where,

E = Electric Field = ?

m = Mass of the Charge = 5 x 10⁻³ kg

g = acceleration due to gravity = 9.8 m/s²

q = magnitude of charge = 4 μC = 4 x 10⁻⁶ C

Therefore,

E(4 x 10⁻⁶ C) = (5 x 10⁻³ kg)(9.8 m/s²)

E = 0.049 N/4 x 10⁻⁶ C

E = 12.25 x 10³ N/C = 12.25 KN/C

Answer:

Acidosis is defined as the formation of excessive acid in the body due to kidney disease or kidney failure.

In order to compensate acidosis, the kidneys will reabsorb more HCO3 from the tubular fluid through tubular cells and collecting duct cell will secret more H+ and ammoniagenesis, which form more NH3 buffer.

Answer:

20J

Explanation:

The computation of the change in kinetic energy is shown below:

As we know that

Work was done on system =  change in potential energy + change in kinetic energy + change in thermal energy

-430J = -510J + change in kinetic energy + 100J

-430J = -410J + change in kinetic energy

So, the change in kinetic energy is 20J

We simply applied the above formula to find out the change in kinetic energy

Answer:

the answer is c

Explanation:

the balloon is lighter than the air around displaced by it

Answer:

0

Explanation:

On a graph, if there is plot of time vs velocity, then area of velocity plot gives the displacement.

also we can see area of plot is velocity* time which is equal to formula of displacement.

area for this plot is velocity * time

Thus,

from

t =0 to t = 0.2

v = 20

t = 0.2 - 0 = 0.2

thus, displacement till 0.2 seconds = 20*0.2 = 4 Km

____________________________________________________

from

t =0 to t = 0.4

v = 0

t = 0.4 - 0.2 = 0.2

thus, displacement from  0.2 seconds to 0.4 seconds = 0*0.2 = 0 Km

____________________________________________________

from

t =0.4 to t = 0.6

v = -20

t = 0.6 - 0.4 = 0.2

thus, displacement from  0.4 seconds to 0.6 seconds = -20*0.2 = -4 Km

Thus, total displacement = 4+0 -4 = 0

Thus, net displacement made by car is 0.

Answer:

An integral or whole multiple of the wavelength λ

or d sin θ = mλ,

for m = 0, 1, −1, 2, −2, ...,

Explanation:

In the double slit interference pattern, if we consider how two waves travel from the slits to the screen, we'll see that each slit is a different distance from a given point on the screen hence, they posses different wavelengths. Waves in a double slit experiment will be in phase if they interfere constructively by starting out crest to crest, or trough to trough. If the waves arrive crest to trough, they will interfere destructively, and arrive out of phase. A constructive interference occurs when the path length difference of the waves exiting the two slits forms an integral multiple of wavelength at the screen. A destructive interference occurs if the path length  differs by half a wavelength. Constructive interference forms the bright fringes, while the dark fringes are formed by destructive interference.

Answer:

The answer is below

Explanation:

Given that:

The area of the plates is 6 m by 0.030 m, Therefore the area = 6 m × 0.03 m =  0.18 m²

the relative permittivity of dielectric (εr) is 7.0

Permittivity of free space (εo) = 8.854 × 10^(-12)

capacitance of 100uF

potential difference (V) of 12V

d = separation between plate

The capacitance (C) of a capacitor is given by:

[tex]C=\frac{\epsilon_o \epsilon_r A }{d}\\ 100*10^{-6}=\frac{8.854*10^{-12}*7*0.18}{d}\\ d=\frac{8.854*10^{-12}*7*0.18}{100*10^{-6}}=1.11*10^{-7}\ m[/tex]

The electric field between plates is given as:

E = V /d

[tex]E = 12 / 1.11*10^{-7}=10.75*10^7\ V/m[/tex]