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Sliva [168]
3 years ago
15

Four charges of equal magnitude q = 2.16 µC are situated as shown in the diagram below. If d = 0.88 m, find the electric potenti

al at location A due to the four charges. (enter the magnitude only)

Physics
1 answer:
Triss [41]3 years ago
8 0

Answer:

<em>The total potential (magnitude only) is 11045.45 V</em>

Explanation:

<u>Electric Potential </u>

The total electric potential at location A is the sum of all four individual potentials produced by the charges, including the sign since the potential is a scalar magnitude that can be computed by

\displaystyle V=\frac{kq}{r}

Where k is the Coulomb's constant, q is the charge, and r is the distance from the charge. Let's find the potential of the rightmost charge:

\displaystyle V_1=\frac{9\cdot 10^{9}\times -2.16\cdot 10^{-6}}{0.88}=-22090.91\ V

The potential of the leftmost charge is exactly the same as the above because the charges and distances are identical

V_2=-22090.91\ V

The potential of the topmost charge is almost equal to the above computed, is only different in the sign:

V_3=+22090.91\ V

The bottom charge has double distance and the same charge, thus the potential's magnitude is half the others':

\displaystyle V_4=\frac{9\cdot 10^{9}\times 2.16\cdot 10^{-6}}{1.76}=+11045.45 \ V

The total electric potential in A is

V=-22090.91\ V-22090.91\ V+22090.91\ V+11045.45 \ V

V=-11045.45 \ V

The total potential (magnitude only) is 11045.45 V

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