1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Pavel [41]
4 years ago
7

The electric field midway between two equal but opposite point charges is 386 n/c, and the distance between the charges is 16.0

cm. what is the magnitude of the charge on each?
Physics
1 answer:
Ronch [10]4 years ago
4 0
<span>We'll use the following formular E = Kq / r^2 Since the electric field is 386 N/C midway between the charges, r = 1/2 * 16 = 8 cm = 0.08m. So we have 386 = (8.998 * 10^(9) * q) / (0.08)^2 386 * 0.0064 = 8.998 * 10^(9) * q q = (386 * 0.0064)/ 8.998 * 10^(9) q = 2.4704/ 8.998 * 10^(9) q = 2.754 * 10^(-10)</span>
You might be interested in
A point charge with a charge q1 = 2.30 μC is held stationary at the origin. A second point charge with a charge q2 = -5.00 μC mo
Alla [95]

Answer:

W = 2.74 J

Explanation:

The work done by the charge on the origin to the moving charge is equal to the difference in the potential energy of the charges.

This is the electrostatic equivalent of the work-energy theorem.

W = \Delta U = U_2 - U_1

where the potential energy is defined as follows

U = \frac{1}{4\pi\epsilon_0}\frac{q_1q_2}{r^2}

Let's first calculate the distance 'r' for both positions.

r_1 = \sqrt{(x_1 - x_0)^2 + (y_1 - y_0)^2} = \sqrt{(0.170 - 0)^2 + (0 - 0)^2} = 0.170~m\\r_2 = \sqrt{(x_2 - x_0)^2 + (y_2 - y_0)^2} = \sqrt{(0.250 - 0)^2 + (0.250 - 0)^2} = 0.353~m

Now, we can calculate the potential energies for both positions.

U_1 = \frac{kq_1q_2}{r_1^2} = \frac{(8.99\times 10^9)(2.3\times 10^{-6})(-5\times 10^{-6})}{(0.170)^2} = -3.57~J\\U_2 = \frac{kq_1q_2}{r_2^2} = \frac{(8.99\times 10^9)(2.3\times 10^{-6})(-5\times 10^{-6})}{(0.3530)^2} = -0.829~J

Finally, the total work done on the moving particle can be calculated.

W = U_2 - U_1 = (-0.829) - (-3.57) = 2.74~J

4 0
3 years ago
Read 2 more answers
One of the main differences between the intaglio and the relief printing processes is that with intaglio the ink ________ the su
TEA [102]
The answer to this question is "LIES BELOW THE SURFACE" happens or occurs. When one of the main differences between the two which is the Intaglio and the other one is the relief printing processes is that with the Intaglio the ink LIES BELOW the surface of the printing plate.
5 0
4 years ago
What was one of the main benefits to the use of the icebox for cooling food? Select all that apply. provided a source for cold w
cupoosta [38]

The correct answers are:

  • It slowed microbial growth in food.
  • It allowed people to store a few days of food in their homes.

Hope I helped! If so, please feel free to rate my answer and consider giving it the Brainliest.


4 0
3 years ago
Read 2 more answers
Is it proper to use an infinitely long cylinder model when finding the temperatures near the bottom or top surfaces of a cylinde
Gelneren [198K]

Answer:

No, it is not proper to use an infinitely long cylinder model when finding the temperatures near the bottom or top surfaces of a cylinder.

Explanation:

A cylinder is said to be infinitely long when is of a sufficient length. Also, when the diameter of the cylinder is relatively small compared to the length, it is called infinitely long cylinder.

Cylindrical rods can also be treated as infinitely long when dealing with heat transfers at locations far from the top or bottom surfaces. However, it not proper to treat the cylinder as being infinitely long when:

* When the diameter and length are comparable (i.e have the same measurement)

When finding the temperatures near the bottom or top of a cylinder, it is NOT PROPER TO USE AN INFINITELY LONG CYLINDER because heat transfer at those locations can be two-dimensional.

Therefore, the answer to the question is NO, since it is not proper to use an infinitely long cylinder when finding temperatures near the bottom or top of a cylinder.

8 0
3 years ago
A school bus moves at speed of 35 mi/hr for 20 miles. How long will it take the bus to get to school
ludmilkaskok [199]

Answer:

Explanation:

Assuming school is at the end of the 20 mile route, then

20 mi / 35 mi/hr = 0.57142...hr

which is about 34 minutes 17 seconds

6 0
3 years ago
Other questions:
  • When you set something down on the ground what kind of work is your arm doing
    10·2 answers
  • You are climbing in the High Sierra where you suddenly find yourself at the edge of a fog-shrouded cliff. To find the height of
    6·1 answer
  • What energy roles do organisms play in an ecosystem?
    15·1 answer
  • The ultracentrifuge is an important tool for separating and analyzing proteins. Because of the enormous centripetal acceleration
    10·1 answer
  • Going around a ferris wheel without stopping is an example of _____ circular motion<br>(apex)
    11·2 answers
  • Which letter below represents a nucleotide?<br> A.<br> D.<br> B.<br> C
    5·2 answers
  • pressure A vertical piston cylinder device contains a gas at an unknown pressure. If the outside pressure is 100 kPa, determine
    5·1 answer
  • Identify the parts of the wave below. Please help! I need it quite soon!
    14·1 answer
  • On a spaceship designed to support a multiyear voyage to the outer planets of the solar system, plants will be grown to provide
    14·1 answer
  • A bus manufacturer decides to double a window's area in order to give passengers a
    14·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!