PHYSICS COLLEGE

Points A, B, and C are at the corners of an equilateral triangle of side 8 m. Equal positive charges of 4 mu or micro CC are at A and B. (a) What is the potential at point C? 8.990 kV * [2.5 points] 2 attempt(s) made (maximum allowed for credit = 5) [after that, multiply credit by 0.5 up to 10 attempts] 8.990 OK (b) How much work is required to bring a positive charge of 5 mu or micro CC from infinity to point C if the other charges are held fixed? .04495 J * [2.5 points] 1 attempt(s) made (maximum allowed for credit = 5) [after that, multiply credit by 0.5 up to 10 attempts] .04495 OK (c) Answer parts (a) and (b) if the charge at B is replaced by a charge of -4 mu or micro CC. Vc= kV [2.5 points] 0 attempt(s) made (maximum allowed for credit = 5) [after that, multiply credit by 0.5 up to 10 attempts] W =

Answers

Answer 1

Answer:

Answer:

a) 8.99*10³ V b) 4.5*10⁻² J c) 0 d) 0

Explanation:

The electrostatic potential V, is the work done per unit charge, by the electrostatic force, producing a displacement d from infinity (assumed to be the reference zero level).
For a point charge, it can be expressed as follows:

$V =(k*q)/(d)$

As the electrostatic force is linear with the charge (it is raised to first power), we can apply superposition principle.
This means that the total potential at a given point, is just the sum of the individual potentials due to the different charges, as if the others were not there.
In our case, due to symmetry, the potential, at any corner of the triangle, is just the double of the potential due to the charge located at any other corner, as follows:

$V = (2*q*k)/(d) = (2*8.99e9N*m2/C2*4e-6C)/(8m) =\n \n V= 8.99e3 V$

The potential at point C is 8.99*10³ V

The work required to bring a positive charge of 5μC from infinity to the point C, is just the product of the potential at this point times the charge, as follows:

$W = V * q = 8.99e3 V* 5e-6C = 4.5e-2 J$

The work needed is 0.045 J.

If we replace one of the charges creating the potential at the point C, by one of the same magnitude, but opposite sign, we will have the following equation:

$V = (8.99e9N*m2/C2*(4e-6C))/(8m) + ((8.99e9N*m2/C2*(-4e-6C))/(8m)) = 0$

This means that the potential due to both charges is 0, at point C.

If the potential at point C is 0, assuming that at infinity V=0 also, we conclude that there is no work required to bring the charge of 5μC from infinity to the point C, as no potential difference exists between both points.

Related Questions

A 50n brick is suspended by a light string from a 30kg pulley which may be considered a solid disk with radius 2.0m. the brick is released from rest and falls to the floor below as the pulley rotates. it takes 4 seconds for the brick to hit the floor. i) what is the tension in newtons in the string well the brick is falling? ii) what is the magnitude of the angular momentum in kg*m^2/s of the pulley at the instant the brick hits the floor?
A wooden block with mass 1.05 kg is placed against a compressed spring at the bottom of a slope inclined at an angle of 35.0 degrees (point A). When the spring is released, it projects the block up the incline. At point B, a distance of 4.90m up the incline from A, the block is moving up the incline at a speed of 5.10 m/s and is no longer in contact with the spring. The coefficient of kinetic friction between the block and incline is 0.55. The mass of the spring is negligible. Calculate the amount of potential energy that was initially stored in the spring.
Susan’s 10 kg baby brother Paul sits on a mat. Susan pulls the mat across the floor using a rope that is angled 30° above the floor. The tension is a constant 30 N and the coefficient of friction is 0.20. Use work and energy to find Paul’s speed after being pulled 3.0 m.
Write down the DE of motion of a particle moving under the influence of gravity and experiencing a resistive force. .
A car is initially moving at 35 km/h along a straight highway. To pass another car, it speeds up to 135 km/h in 10.5 seconds at a constant acceleration.(a) how large was the acceleration in m/s ^2(b)how large was the acceleration, in units go g= 9.80 m/s ^2

A wire is wrapped around a piece of iron, and then electricity is run through the wire. What happens to the iron?

Answers

Search ResultsBy simply wrapping wire that has an electrical current running through it around a nail, you can make an electromagnet. When the electric current moves through a wire, it makes a magnetic field. ... You can make a temporary magnet by stroking apiece of iron or steel (such as a needle) along with a permanent magnet.

Hope This Helps!

We showed that the length of the pendulum of period 2.000 seconds on the Earth’s surface was 0.99396 meters. What period would this same pendulum have on the surface of Mars? What length would the pendulum be in order to have a period of 2.000 seconds?

Answers

To solve this problem it is necessary to apply the concepts related to the Period based on gravity and length.

Mathematically this concept can be expressed as

$T= 2\pi \sqrt{(l)/(g)}$

Where,

l = Length

g = Gravitational acceleration

First we will find the period that with the characteristics presented can be given on Mars and then we can find the length of the pendulum at the desired time.

The period on Mars with the given length of 0.99396m and the gravity of the moon (approximately $1.62m / s ^ 2)$ will be

$T= 2\pi \sqrt{(l)/(g)}$

$T= 2\pi \sqrt{(0.99396)/(1.62)}$

$T = 4.921seg$

For the second question posed, it would be to find the length so that the period is 2 seconds, that is:

$T= 2\pi \sqrt{(l)/(g)}$

$2= 2\pi \sqrt{(l)/(1.62)}$

$l = 0.16414m$

Therefore, we can observe also that the shorter distance would be the period compared to the first result given.

The volume control on a stereo is designed so that three clicks of the dial increase the output by 10 dB. How many clicks are required to increase the power output of the loudspeakers by a factor of 100?

Answers

Answer:

300 clicks...

Explanation:

Output on 3 clicks = 10 dB

Increasing 10 by a factor of 100 equals 1000 dB so,

Its simple math, clicks will also increase in the same ratio and it shall take 300 clicks to increase the volume by a factor of 100.

Consider a bird that flies at an average speed of 10.7 m/s and releases energy from its body fat reserves at an average rate of 3.70 W (this rate represents the power consumption of the bird). Assume that the bird consumes 4.00g of fat to fly over a distance db without stopping for feeding. How far will the bird fly before feeding again?Fat is a good form of energy storage because it provides the most energy per unit mass: 1.00 grams of fat provides about 9.40 (food) Calories, compared to 4.20 (food) Calories per 1.00 grams of carbohydrate. Remember that Calories associated with food, which are always capitalized, are not exactly the same as calories used in physics or chemistry, even though they have the same name. More specifically, one food Calorie is equal to 1000 calories of mechanical work or 4186 joules. Therefore, in this problem use the conversion factor 1Cal=4186J.

Answers

Answer:

The distance covered by the bird before feeding is $4.55 * 10^(5)$ m.

Explanation:

As the bird consumes 4 g of fat before flying, the amount of initial food energy ( $E_(F)$ ) stored by it is given by

$E_(F) = 4 g * 9.4 (food) cal = 37.6 (food) cal$

So the mechanical energy stored by the bird ( $E_(M)$ ) is given by

$E_(M) = E_(F) * 4186 J = 1.57 * 10^(5) J$

Given, the power consumed by the bird $P = 3.7 W$

So, the time ( $t$ ) required to consume this power by the bird is

$t = (E_(M))/(P) = (1.57 * 10^(5) J)/(3.7 W) = 4.24 * 10 ^(4) s$

As the bird flies at an average speed ( $v$ ) of $10.7 ms^(-1)$ , so the distance ( $d$ ) covered by the bird before feeding again is given by

$d = v * t = 10.7 ms^(-1) * 4.25 * 10 ^(4) s = 4.55 * * 10^(5) m$

The distance of the bird'sflight before him/her feeds again is mathematically given as

d = 4.55* 10^{5} m

What is the distance of the bird's flight before him/her feeds again?

Question Parameter(s):

a bird that flies at an averagespeed of 10.7 m/s

its body fat reserves at an average rate of 3.70 W

the most energy per unit mass: 1.00 grams of fat provides about 9.40 (food) Calories,

Generally, the initial food energy is mathematically given as

Ex= 4 g*9.4

Ex= 37.6cal

Therefore, the mechanical energy

Em = Ex * 4186

Em = 1.57*10^{5} J

In conclusion, time of flight

$t = (E_(M))/(P) \n\n t=(1.57 *10^(5) J)/(3.7 W) \n$

t= 4.24*10 ^{4} s

Th distance hence is

d = v* t

d= 10.7 *4.25*10 ^{4}

d = 4.55* 10^{5} m

Answers

Answer:

The environment is warmed by the light throughout the day, such that the temperature increases. The weather is decreasing and the temperature decreases in the night as the sun falls. There was a misunderstanding. Thanks to the density, the atmosphere becomes densest on the earth. The air becomes colder and colder when you move up.

Explanation:

Answer is above

Hope this helps.

The universe is filled with photons left over from the Big Bang that today have an average energy of about 2 × 10−4 eV (corresponding to a temperature of 2.7 K). As derived in lecture, the number of available energy states per unit volume for photons is ????(????)????????