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Anettt [7]
3 years ago
12

A charge of 3 nC is uniformly distributed around a ring of radius 10 cm that has its center at the origin and its axis along the

x axis. A point charge of 1 nC is located at x = 2.5 m. Find the work required to move the point charge to the origin. Hint: The electric field on the axis was calculated in class (see Example 3 in the the slides for Lecture 16).
Physics
1 answer:
Step2247 [10]3 years ago
8 0

Answer:

W=2.592*10^{-7} J

Explanation:

GIVEN DATA:

Charge 3 nC

Radius of ring is 10 cm

A point charge 1 nC

Distance of point charge is 2.5 m

We know that voltage is calculated as

V(0) =\frac{KQ}{r}

      =\frac{(9*10^{-9})(3*10^{-9})}{0.1}

V = 270 V

At x = 2.5m

r = \sqrt{(0.1^2+2.5^2)}

r = 1.581 m

V(2) = \frac{(9*10^9)(3*10^{-9})}{1.581}

V(2) = 10.801 V

Work done is calculated as

W= q(dV)

W = (1*10^{-9})(270 - 10.80)

W=2.592*10^{-7} J

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Explanation:

a.)

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b.)

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By the definition of Acceleration (a)

             a = \frac{v_{2} - v_{1} } {t}  

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c.)

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             a = \frac{v_{2} - v_{1} } {t}  

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