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lana66690 [7]
3 years ago
8

Two rings of radius 5 cm are 20 cm apart and concentric with a common horizontal x-axis. The ring on the left carries a uniforml

y distributed charge of +30 nC, and the ring on the right carries a uniformly distributed charge of −30 nC. What is the electric field due to the right ring at a location midway between the two rings?
Physics
1 answer:
Yanka [14]3 years ago
3 0

Answer:

The electric field due to the right ring at a location midway between the two rings is 2.41\times10^{3}\ V/m

Explanation:

Given that,

Radius of first ring = 5 cm

Radius of second ring = 20 cm

Charge on the left of the ring = +30 nC

Charge on the right of the ring = -30 nC

We need to calculate the electric field due to the right ring at a location midway between the two rings

Using formula of  electric field

E=\dfrac{1}{4\pi\epsilon_{0}}\times\dfrac{qx}{(x^2+R^2)^{\frac{3}{2}}}

Put the value into the formula

E=\dfrac{9\times10^{9}\times30\times10^{-9}\times0.1}{((0.1)^2+(0.2)^2)^{\frac{3}{2}}}

E=2.41\times10^{3}\ V/m

Hence, The electric field due to the right ring at a location midway between the two rings is 2.41\times10^{3}\ V/m

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Answer:

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Explanation:

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now by Snell'a law at that interface we have

\mu_1 sin\theta_i = \mu_2 sin \theta_r

now we will have

1.52  sin35.5 = 1.40 sin\theta_r

\theta_r = 39.12^0

now this is the angle of incidence for oil air interface

so now again by Snell's law we will have

\mu_2 sin\theta_i' = \mu_{air} sin\theta

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In this situation,

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r = 4.60 cm = 0.046 m is the distance of the electron from the wire

Therefore the magnetic field strength at the electron's location is

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The force exerted on a charged moving particle travelling perpendicular to a magnetic field is given by

F=qvB

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q is the magnitude of the charge of the particle

v is its velocity

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For the electron of this problem,

q=1.6\cdot 10^{-19} C is the charge

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Now the direction of the force can be found by using the right-hand rule. We have:

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- middle finger = direction of the magnetic field (to the north)

- Thumb = direction of the force --> downward

However, the electron carries a negative charge, so we must reverse the direction of the force: therefore, the force experienced by the electron will be upward, so in the same direction as the current in the wire.

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