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Two charges +6.0 mu C and -2.0 mu C are fixed at p

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Answer #1

a) electric potential at point C = kq1/r1 + kq2/r2

= 9*10^9 *10^-6 * (6 / sqrt(0.5^2 + 0.4^2) - 2 /0.5 )

= 48333.83 V

electric potential at D = 9*10^9 *10^-6 * (6 / 0.7 - 2 /0.3 ) = 17142..85 V

b) work done in moving a charge from D to C

= q (Vc - Vd)

= 2*10^-9*(48333.83 - 17142.85)

= 6.24*10^-5 J

c) the work is converted to potential energy of the charge

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