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d
Explanation: V = (1/4πεo) ∑Q/r = (10 X 10-9
/4πεo)
(0.5 + 0.33 + 0.25 + 0.2 + 0.166 + 0.142) = 143.35 volts.
To find the electric potential at the origin due to six point charges, (Q = 10 , text{nC} = 10 times 10^{-9} , text{C}), located at distances of 2, 3, 4, 5, 6, and 7m, we can use the formula:
[V = frac{1}{4piepsilon_0} sum frac{Q}{r_i}]
Where:
– (V) is the electric potential,
– (Q) is the charge,
– (r_i) is the distance of each charge from the point where the potential is being calculated (in this case, the origin),
– (epsilon_0) is the permittivity of free space ((8.85 times 10^{-12} , text{F/m})).
Plugging in the values:
[V = frac{1}{4pi(8.85 times 10^{-12})} left( frac{10 times 10^{-9}}{2} + frac{10 times 10^{-9}}{3} + frac{10 times 10^{-9}}{4} + frac{10 times 10^{-9}}{5} + frac{10 times 10^{-9}}{6} + frac{10 times 10