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A positive point charge (q = +8.65 x 10-8 C) is surrounded by an equipotential surface A, which has a radius of rA = 2.97 m. A positive test charge (q0 = +4.56 x 10-11 C) moves from surface A to another equipotential surface B, which has a radius rB. The work done by the electric force as the test charge moves from surface A to surface B is WAB = -6.95 x 10-9 J. Find rB.

Respuesta :

Answer:

[tex]r_B = 1.88 m[/tex]

Explanation:

As we know that work done by electric force is given as

[tex]W_e = -q\Delta V[/tex]

so here we know that charge is moving from

[tex]r_A = 2.97 m[/tex]

to another position

so we will have

[tex]W_{AB} = \frac{kq_1q_2}{r_A} - \frac{kq_1q_2}{r_B}[/tex]

[tex]-6.95\times 10^{-9} = (9\times 10^9)(8.65\times 10^{-8})(4.56\times 10^{-11})(\frac{1}{2.97} - \frac{1}{r_B})[/tex]

[tex]-0.196 = (\frac{1}{2.97} - \frac{1}{r_B})[/tex]

[tex]r_B = 1.88 m[/tex]

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