Respuesta :
The Potential at the origin due to the four point charges located at (0, 2), (0, -2), (2 , 0), and (-2 , 0) will be - [tex]45\times10^{4}[/tex] volts.
We have four point charges, each of charge [tex]2.5\times10^{-5}[/tex] C, are located on the x - axis and y - axis, one at each of the locations (0, 2), (0, -2), (2 , 0), and (-2 , 0).
We have to calculate the potential at the origin O(0,0).
What is the formula to calculate the Electric potential at a point ' r ' meters away from a point charge ' q ' ?
The Electric field potential due to point charge ' q ' at a distance ' r ' meters apart is -
[tex]V(r) = \frac{1}{4\pi\epsilon_{o} } \frac{q}{r}[/tex]
Now, in the question given to us -
q = [tex]2.5\times10^{-5}[/tex]
Let us label the four points as - A (0, 2), B (0, -2), C (2 , 0), and D (-2 , 0).
Now, the potential due to charge located at point A will be -
[tex]V_{A} = \frac{1}{4\pi\epsilon_{o} } \frac{q}{r}\\V_{A} = \frac{1}{4\pi\epsilon_{o} } \frac{q}{r}\\V_{A} = 9\times10^{9}\times\frac{2.5\times10^{-5} }{2} \\V_{A} = 11.25\times10^{4}[/tex]
Now, the distance of the center from each point will be same = 2 m.
Hence, the potential due to each charge at the origin will be -
[tex]V_{A} =V_{B} =V_{C} =V_{D} = 11.25\times10^{4}[/tex]
Since, the potential is a scalar quantity, we can add the potential due to each charge using simple addition -
[tex]V_{A} +V_{B} +V_{C} +V_{D} = 4\times11.25\times= 45\times10^{4}[/tex] Volts
Hence, the potential at the center will be - [tex]45\times10^{4}[/tex] volts.
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