The potential at point A (relative to infinity) due to the given charges is 89900V
Given the data in the question.
As illustrated in the diagram, the distance between the first charge [tex]q_1[/tex] and point A is 0.800m and distance between the second charge [tex]q_2[/tex] and point A is 0.400m.
So,
[tex]q_1[/tex] = +2.00 μC = 2 × 10⁻⁶ C [Convert Microcoulombs (μC) to coulombs (C)]
[tex]q_2[/tex] = +4.00 μC = 4 × 10⁻⁶ C [Convert Microcoulombs (μC) to coulombs (C)]
[tex]r_1[/tex] = 0.400m
[tex]r_2[/tex] = 0.800m
In the question also, [tex]k = \frac{1}{4\pi \epsilon _0} = 8.99 *10^9N.m^2/C^2[/tex]
Now, We determine the potential at point A (relative to infinity) due to given charges.
Potential at Point A is the sum of the potentials due to each of the charges individually.
From Coulomb's Law
[tex]V_A = \frac{1}{4\pi \epsilon _0} [ \frac{q_1}{r_1} + \frac{q_2}{r_2} ]}[/tex]
Since [tex]k = \frac{1}{4\pi \epsilon _0}[/tex]
[tex]V_A = k [ \frac{q_1}{r_1} + \frac{q_2}{r_2} ]}[/tex]
We substitute our given values into the formula
[tex]V_A = (8.99 * 10^9 N.m^2/C^2)[\frac{2*10^{-6}C}{0.400m} + \frac{4*10^{-6}C}{0.800m}][/tex]
[tex]V_A = 89900V[/tex]
Therefore, the potential at point A (relative to infinity) due to the given charges is 89900V
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