Physics, 27.09.2019 03:30, texassaint797

Acharge q is distributed uniformly around the perimeter of a ring of radius r. determine the electric potential difference between the point at the center of the ring and a point on its axis at a distance 6r from the center. (use any variable or symbol stated above along with the following as necessary: ke.) δv

Answers: 1

Show answers

Answers

answered: geraldmorgan5580

$\rm 0.8356\ \dfrac{kQ}{R}.$

Explanation:

The electric potential due to a charge q at a point r distance arway from it is given by

$\rm V = \dfrac{kq}{r}.$

k is called the Coulomb's constant.

For the given ring of radius R, consider a small element of length dl having charge dq, the electric potential due to this element at a distance r is given by

$\rm dV = \dfrac{k\ dq}{r}.$

The length of the whole ring is equal to the circumference of the ring, $\rm 2\pi R.$

The charge Q is uniformly distributed around the perimeter of the ring therefore, the charge on the element of the length dl, $\rm dq = \dfrac{Q}{2\pi R}\times dl=\dfrac{Q\ dl}{2\pi R}.$

The electric potential at the center of the ring:

The electric potential at the center of the ring due to the small element of ring is given by

$\rm dV_1 = \dfrac{k\ dq}{R}=\dfrac{k}{R}\ \dfrac{Q\ dl}{2\pi R}=\dfrac{kQ\ dl}{2\pi R^2}.$

The electric potential at the center of the ring due to the whole ring is given by

$\rm V_1 = \int dV_1\\=\int\limits^{l = 2\pi R}_ 0\dfrac{kQ\ dl}{2\pi R^2}\\=\dfrac{kQ}{2\pi R^2}\int\limits^{l = 2\pi R}_ 0\ dl\\=\dfrac{kQ}{2\pi R^2}\ 2\pi R\\=\dfrac{kQ}{R}.$

The electric potential at the distance 6R from the center of the ring along its axis:

The distance of the point, which is at 6R distance from the center of the ring, from the small element is $\rm \sqrt{R^2+(6R)^2}=R\sqrt{37}.$

The electric potential at this point due to the small element is given by

$\rm dV_2 = \dfrac{kdq}{R\sqrt{37}}=\dfrac{k}{R\sqrt{37}}\dfrac{Q\ dl}{2\pi R}=\dfrac{kQ\ dl}{2\pi R^2\sqrt {37}}.$

The electric potential at this point due to the whole ring is given by

$\rm V_2 = \int dV_2\\=\int\limits^{l = 2\pi R}_ 0\dfrac{kQ\ dl}{2\pi R^2\sqrt{37}}\\=\dfrac{kQ}{2\pi R^2\sqrt{37}}\int\limits^{l = 2\pi R}_ 0\ dl\\=\dfrac{kQ}{2\pi R^2\sqrt{37}}\ 2\pi R\\=\dfrac{kQ}{R\sqrt{37}}.$

Thus, the electric potential difference between the point at the center of the ring and a point on its axis at a distance 6R from the center is given by

$\rm \Delta V = V_1-V_2\\=\dfrac{kQ}{R}-\dfrac{kQ}{R\sqrt {37}}\\=\dfrac{kQ}{R}\left ( 1-\dfrac{1}{\sqrt{37}} \right )\\=0.8356\ \dfrac{kQ}{R}.$

answered: Guest

can u provide the graph and if its not possible

then i can tell u the segment where the

average time interval starts degrading is the segment where the car is slowing down

answered: Guest

Answerthe energy required is 1.60 × 10⁻¹⁸ j, to three significant figures.explanationwe define electric potential as the ratio electric potential energy to the charge, $v = \dfrac{e_p}{q}$ then the following gives us the potential difference, $\delta v = v_b - v_a$ $\begin{aligned} \delta v & = \dfrac{e_p_b}{q} - \dfrac{e_p_a}{q} \\ & = \dfrac{\delta e_p}{q} \end{aligned}$ the formula $\delta v = \delta e_p / q$ may be on your formula sheet.the change in electric potential energy would be the energy that would needed to move it from point a to point b. solving for that, we get $\delta e_p = q\delta v$ using q = 8.00 × 10⁻¹⁹ c and δv = 2.00 v, we get $\begin{aligned} \delta e_p & = (8.00\times10^{-19} \text{ c})(2.00\text{ v}) \\ & = 1.60\times 10^{-18}\text{ j} \end{aligned}$ the energy required is 1.60 × 10⁻¹⁸ j, to three significant figures.

Other questions on the subject: Physics

Physics, 22.06.2019 16:00, slrich21

While the change in blank will remain the same during a collision, the force needed to bring an object to a stop can be blank if the time if collision is blank

Answers: 1

continue

Physics, 22.06.2019 18:30, lilypup004

Asled is being held at rest on a slope that makes an angle θ with the horizontal. after the sled is released, it slidesa distance d1 down the slope and then covers the distance d2 alongthe horizontal terrain before stopping. find the coefficient ofkinetic friction μk between the sled and the ground, assumingthat it is constant throughout the trip. find the coefficient of kinetic frictionμk. express your answer in terms of someor all of the variables d1, d2, and θ.

Answers: 1

continue

Physics, 22.06.2019 19:00, sanchez626

Friction removes energy from objects in motion. which statement best describes how this works? a) friction transforms ke into thermal energy b) friction transfers thermal energy to ke c) friction transforms te into pe d) friction transforms pe into ke e) friction transfers ke into pe

Answers: 1

continue

Physics, 22.06.2019 22:40, laskew37221

Calculate the volume of this regular solid. what is the volume of the cone? round your answer to the nearest hundredth

Answers: 1

continue

Do you know the correct answer?

Acharge q is distributed uniformly around the perimeter of a ring of radius r. determine the electri...