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Katen [24]
3 years ago
12

(b) Find a point between the two charges on the horizontal line where the electric potential is zero. (Enter your answer as meas

ured from q1.)
Physics
2 answers:
tatuchka [14]3 years ago
8 0

Answer:

The question seems to be incomplete, but ill try to answer it the best that I can do it.

We have two charges in a horizontal line, where the charge of the charges is q1 and q2, and the distance in between the charges is d. So we can think that one of the charges is in x = 0 and the other charge is in x = d, and we can solve it only on x.

We want to find a point in between the charges where the electric potential is zero.

The electric potential for a charge is:

V(x) = k*q/Ix - x0I

where x0 is the position of the charge, and k is a constant.

Then the potentials that we have aer

V1(x) = k*q1/Ix-0I = k*q1/IxI

V2(x) =  k*q1/IxI

now we want to find for wich value of x V1(x) + V2(x) = 0

k*q1/IxI +  k*q2/Ix - dI = 0

k(q1/IxI +  q2/Ix - dI) = 0  

So here we can see that the signs of q1 and q2 must be different:

q1/IxI = -q2/Ix-dI

q1*Ix-dI = -q2*IxI

With this relation, you can find x.

I can not keep solving it without the values of q1, q2, and d, but with this equation, you can find the value of x.

aleksley [76]3 years ago
3 0

Complete Question: A charge q1 = 2.2 uC is at a distance d= 1.63m from a second charge q2= -5.67 uC. (b) Find a point between the two charges on the horizontal line where the electric potential is zero. (Enter your answer as measured from q1.)

Answer:

d= 0.46 m

Explanation:

The electric potential is defined as the work needed, per unit charge, to bring a positive test charge from infinity to the point of interest.

For a point charge, the electric potential, at a distance r from it, according to Coulomb´s Law and the definition of potential, can be expressed as follows:

V = \frac{k*q}{r}

We have two charges, q₁ and q₂, and we need to find a point between them, where the electric potential due to them, be zero.

If we call x to the distance from q₁, the distance from q₂, will be the distance between both charges, minus x.

So, we can find the value of x, adding the potentials due to q₁ and q₂, in such a way that both add to zero:

V = \frac{k*q1}{x} +\frac{k*q2}{(1.63m-x)} = 0

⇒k*q1* (1.63m - x) = -k*q2*x:

Replacing by the values of q1, q2, and k, and solving for x, we get:

⇒ x = (2.22 μC* 1.63 m) / 7.89 μC = 0.46 m from q1.

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Compare and contrast series and parallel circuits?
dalvyx [7]

In a series circuit, a common current flows through all the components of the circuit. While in a parallel circuit, a different amount of current flows through each parallel branch of the circuit. Whereas in the parallel circuit, the same voltage exists across the multiple components in the circuit.

Hope It Helps!

6 0
3 years ago
An object is located 50 cm from a converging lens having a focal length of 15 cm. Which of the following is true regarding the i
crimeas [40]

Answer:

It is real, inverted, and smaller than the object.

Explanation:

Let's start by using the lens equation to find the location of the image:

\frac{1}{q}=\frac{1}{f}-\frac{1}{p}

where we have:

q = ? is the distance of the image from the lens

f = 15 cm is the focal length (positive for a converging lens)

p = 50 cm is the distance of the object from the lens

Solving the equation for q, we find

\frac{1}{q}=\frac{1}{15 cm}-\frac{1}{50 cm}=0.047 cm^{-1}

q=\frac{1}{0.047 cm^{-1}}=+21.3 cm

The sign of q is positive, so the image is real.

Now let's also write the magnification equation:

h_i = - h_o \frac{q}{p}

where  

h_i, h_o are the size of the image and of the object

By substituting p = 50 cm and q = 21.3 cm, we find

h_i = - h_o \frac{21.3 cm}{50 cm}=-0.43 h_o

So we notice that:

|h_i| < |h_o| : this means that the image is smaller than the object

h_i < 0 : this means that the image is inverted

so, the correct option is:

It is real, inverted, and smaller than the object.

7 0
3 years ago
Why it is important to have exact standards of measurement
Tamiku [17]

Answer:

reference against doubt

Explanation:

exact standards of measurement are a reference point for situations in doubt , a line against which to test and observe.

6 0
4 years ago
A shot-putter released the shot at an angle of 41.5 degrees and a height of 1.9 m with an initial velocity of 13.3 m/s. How far
liq [111]

Answer:

x = 17.88[m]

Explanation:

We can find the components of the initial velocity:

(v_{x})_{o}  = 13.3*cos(41.5)=9.96[m/s]\\(v_{y})_{o}  = 13.3*sin(41.5)=8.81[m/s]

We have to remember that the acceleration of gravity will be worked with negative sign, since it acts in the opposite direction to the movement in direction and the projectile upwards.

g = - 9.81[m/s^2]

Now we must find the time it takes for the projectile to hit the ground, as the problem mentions that it does not impact on the board.

y=y_{o} +(v_{y} )_{o} *t-0.5*g*(t)^{2} \\0=1.9+(8.81*t)-(4.905*t^{2})\\-1.9=8.81*t*(1-0.5567*t)\\t=0\\t=1.796[s]

With this time we can calculate the horizontal distance:

x=(v_{x})_{o} *t\\x=9.96*1.796\\x=17.88[m]

3 0
3 years ago
Substances that move to the stronger parts of a magnetic field are termed ______ substances; the atomic feature responsible for
Arisa [49]

Substances that move to the stronger parts of a magnetic field are termed paramagnetic substances; the atomic feature responsible for this property is presence of unpaired electrons in atoms.

<h3>What is a paramagnetic substance?</h3>

A paramagnetic substance is the substance that possess unpaired electrons that are heavily attracted in a magnetic field.

A magnetic field is defined as the field that exists around a magnet that produces a field of force.

Examples of paramagnetic substance include the following:

  • aluminum,
  • gold,
  • copper.
  • Chromium, and
  • Manganese.

These substances are known as paramagnetic substances because they possess a high number of unpaired electrons.

Other properties of a paramagnetic substance include the following:

  • They have a permanent dipole moment or permanent magnetic moment.
  • They are weakly magnetized in the direction of the magnetizing field.
  • They usually have constant relative permeability (μr) slightly greater than 1.

Therefore, Substances that move to the stronger parts of a magnetic field are termed paramagnetic substances; the atomic feature responsible for this property is presence of unpaired electrons in atoms.

Learn more about magnets here:

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3 0
1 year ago
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