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Figure < 2 of 2 92 ity q za di tx 91
Review (Figure 1) Make sure the angle setting in your calculator is "Degree". sin30º = 0.5 Before putting any charge at Origin. The configuration is shown in (Figure 1) 91 = -8.80 C 92 = 2.90 4C1 HC = 10-6C use k = 9.00 x 10' N/(m2 c) for the Coulomb Constant 91 = -8.80 uc q2 = 2.90 C 1 MC = 10-6C use k = 9.00 x 10 N/(m2 c2) for the Coulomb Constant. di = 0.320 m angle a = 26° d2 = 0.260 m = di = 0.320 m 0.260 m Q = 26° d2 = Part A - Calculate the electric POTENTIAL at the origin by charge q1. AZO XX vi i Use scientific notation is Newton/Coulomb N/C, NO NEED to enter unit. answer format: 1.234.10" , unit V1 = -2.475x105 v Submit Previous Answers Correct Important: If you use this answer in later parts, use the full unrounded value in your calculations. Part B - Calculate the electric POTENTIAL at the origin by charge 92. V2 = 1.004x105 v Submit Previous Answers Correct Correct answer is shown. Your answer 100384.62 V was either rounded differently or used a different number of significant figures than required for this part Important: If you use this answer in later parts, use the full unrounded value in your calculations.
m Review Part C - Calculate the TOTAL electric POTENTIAIL at the origin by charge 1 (Figure 1) and q2. Vtotal 91 = -8.80 uc 92 = 2.90 4C 1 HC = 10-6C use k = 9.00 x 10°N/(m2c2) for the Coulomb Constant. -1.471x105 v = dị = 0.320 m 2 = 26 º d2 = 0.260 m Submit Previous Answers Correct Correct answer is shown. Your answer -147115.38 V was either rounded differently or used a different number of significant figures than required for this part. Important: If you use this answer in later parts, use the full unrounded value in your calculations. Now put a Negative point charge q = -1.90 C at origin O. Complete comfiguration is shown in (Figure 2) Part D - Calculate the TOTAL electric POTENTIAL ENERGY (of q). Enter regular number, keep 4 digits after the decimal point. Unit is Joules (no need to enter) EP Etotal 0.2795 J (of a) = Submit Previous Answers Correct Important: If you use this answer in later parts, use the full unrounded value in your calculations.
Review (Figure 1) Submit Previous Answers 91 = -8.80 4C q2 = 2.90 4c 1 AC = 10-6C use k = 9.00 x 10N/(m2 c2 for the Coulomb Constant. Correct Correct answer is shown. Your answer - 147115.38 V was either rounded differently or used a different number of significant figures than required for this part Important: If you use this answer in later parts, use the full unrounded value in your calculations. di = 0.320 m 0.260 m a = 26° d2 = Now put a Negative point charge q = -1.90 vC at origin O. Complete comfiguration is shown in (Figure 2) Part D - Calculate the TOTAL electric POTENTIAL ENERGY (of q). Enter regular number, keep 4 digits after the decimal point. Unit is Joules (no need to enter) EP Etotal 0.2795 J (of a) Submit Previous Answers Correct Important: If you use this answer in later parts, use the full unrounded value in your calculations. Part E-While charges q1 and q2 are held in place, charge q is released from the origin from rest and it moves away. What is its final speed when it reaches the infinity? The mass of charge q is mass = 2.90x10-4 kg Enter a regular number, keep 2 digits after the decimal point. Unit is m/s (no need to enter) Templates @do redo reset keyboard shortcuts help Symbols final speed 0.15 m/s Figure 1 of 2 Submit Previous Answers Request Answer 92 ity x Incorrect: Try Again: 5 attempts remaining