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Answer :
The capacitor should be charged to a potential of approximately 75,317 volts to store 1.70 J of energy.
The energy stored in a capacitor can be calculated using the formula:
E = (1/2) * C * V^2
Where:
E is the energy stored (1.70 J in this case)
C is the capacitance (0.600 μF = 0.600 * 10^(-6) F)
V is the potential (to be determined)
Rearranging the formula, we can solve for V:
V^2 = (2 * E) / C
V = sqrt((2 * E) / C)
V = sqrt((2 * 1.70 J) / (0.600 * 10^(-6) F))
V ≈ sqrt(5.67 * 10^9) V
V ≈ 75,317 V
Therefore, the capacitor should be charged to a potential of approximately 75,317 volts to store 1.70 J of energy.
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