Answer to Question #312490 in Electric Circuits for dan

Question #312490

two charges Q1=+3nC and Q2=-5nC are separated by a distance of 6cm and 2cm for point A amd 2cm and 10cm for point B (a) Calculate the electric potential at point A and point B.



(b) How much work is done by the electric field in a moving 4.00nC particle from point A and point B.

1
Expert's answer
2022-03-16T18:29:41-0400

Potential

VA=kq1rV_A=\frac{kq_1}{r}

VA=9×109×3×1090.022=67500VV_A=\frac{9\times10^9\times3\times10^{-9}}{0.02^2}=67500V

VB=9×109×5×1090.12=4500VV_B=-\frac{9\times10^9\times5\times10^{-9}}{0.1^2}=-4500V

(2)

Work done

W=q(VAVB)W=q(V_A-V_B)


W=4×109×(76500+4500)=3.24×104JW=4\times10^{-9}\times(76500+4500)=3.24\times10^{-4}J


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