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4 years ago

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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).

1 answer:

Step2247 [10]4 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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