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

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Answer from: 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

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

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Answer from: Quest

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