Q3As shown one point charge of 222 pC is located at x500 mm

Q#3:As shown one point charge of -2.22 pC is located at x=-5.00 mm and another point charge of +5.55 pC is located at x=+2.00 mm , both on the x-axis in vacuum. Find the electric potential energy of this system

-2.22 pC +5.55 pC

Q#4: Calculate the electric potential at the origin O (at x=0) due to the point charges of Problem 3 above.

Part of equation used (one EQUAL SIGN)

Solution

#3) q1 = -2.22 pC

q2 = 5.55 pC

distance between charges , d = 5 + 2 = 7 mm

electic potential energy of the system = k * q1 * q2/d^2

electic potential energy of the system = 9 *10^9 * (2.22 *10^-12) * (-5.55 *10^-12)/(0.007)

electic potential energy of the system = -1.584 *10^-11 J

4)

for the electric potential at origin

electric potential at origin = k * q1/d1 + k * q2/d2

electric potential at origin = 9 *10^9 * 1 *10^-15 * (-2.22/0.005 + 5.55/0.002)

electric potential at origin = 0.021 V

the eletric potential at origin is 0.021 V