initially extremely fa apart. How much work does it take to bring the 3.0 C char

Question

initially extremely fa apart. How much work does it take to bring the 3.0 C charge to x = 30 mm, y = 000 mm and the-90-iCharge tox=-3.0 mm, y = 0.00 mm? (1/4th" 899-109 N . 4) (15 points) A-30-uC point charge and a -9.0-C point charge ar A) 40) B)51J C) 27) D) 81 J E) 68] s) 00 point) A cubic surface with 10 cm side is traversed diagonaly by a long uniformly charged wire with linear charge density of 5.5uC/m. The flux through the cube is A) 10,360 Nm2/ )14650 Nm2Ic 107,600 Nm2/C D) 17,940N/C E)87900Nm2/C

Explanation / Answer

potential energy of two point charges is U1 = k*q1*q2/r1

for extremely far apart ,r1 is infinity

then U1 = k*q1*q2/infinity

then U1 = 0 J

for the final points the distnace between the two charges is 6 mm = 6*10^-3 m

then

Now U2 = k*q1*q2/r2 = (8.99*10^9*3*10^-6*9*10^-6)/(6*10^-3) = 40.45 J

then required work done is W = U1 - U2 = 0-40.45 J = 40.45 J

so the answer is A) 40 J

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