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A charge of −9.0 nc is uniformly distributed around a thin plastic ring lying in a yz plane with the ring center at the origin. a −5.6 pc point charge is located on the x axis at x = 2.5 m. for a ring radius of 1.5 m, how much work must an external force do on the point charge to move it to the origin?

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satishsolanki1p34q6g
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okpalawalter8

Work done by an external force is mathematically given as

W=13.99491195J

How much work must an external force do?

Question Parameters:

A charge of −9.0 nc

at the origin. a −5.6 pc point charge is located on the x axis at x = 2.5 m. for a ring radius of 1.5 m,

Generally, the equation for the work  is mathematically given as

W=(v0-Va)q

Therefore

[tex]W=Q/(4\pi e(1/R-1/\sqrt{R^2+x^2}))*q[/tex]

[tex]W=-9*10^{-9}/(4\pi (8.8541878128*10^{-12}*(1/1.5-1/\sqrt{1.5^2+2.5^2}))*-5.6*10^{-2}[/tex]

W=13.99491195J

For more information on Work

https://brainly.com/question/756198

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