Q? = -12.0 µC Q? = +10.0 µC 120 cm +y +x 120 cm 120 cm Q? = -15.0 µC Figure 2 120 cm 80 cm Q? = -5.0 µC 3. From its original position in Figure 1, particle 4 has been moved 80 cm in the positive x direction. How much work ($\Delta PE = PE_{final} - PE_{initial}$) is required to move the charge from its original position (from Figure 1) to the position shown in Figure 2?
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The first statement is "Q = -12.0 C". This means that the charge at this point is -12.0 Coulombs. Show more…
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