The physics of electric potential energy at one corner of a triangle.

Two charges, q1 = +5.0 ìC and q2 = -2.0 ìC are 0.50 m apart. A third charge, q3 = +6.0 ìC is brought from infinity to a spot 0.50 m from both charges so that the three charges form an equilateral triangle.
a) Determine the total electrical potential at the location of q3 created by the charges
q1 and q2.
b) Determine the change in electric potential energy in bringing the charge q3 to this
location.
c) Determine the work required to bring the charge q3 to the location described

To find the total electrical potential at the location of q3, we need to consider the contributions from q1 and q2 separately and then sum them up.

a) The electrical potential at the location of q3 due to q1 can be found using the formula for electric potential:

V1 = k * (q1 / r1),

where k is Coulomb's constant (k = 9.0 x 10^9 N m^2/C^2), q1 is the charge of q1 (+5.0 μC), and r1 is the distance from q1 to q3 (0.50 m).

V1 = (9.0 x 10^9 N m^2/C^2) * (5.0 x 10^-6 C) / 0.50 m.

V1 = 9.0 x 10^9 N m^2/C^2 * 0.010 C / 0.50 m.

V1 = 180 V.

The electrical potential at the location of q3 due to q2 can be found in the same way:

V2 = k * (q2 / r2),

where q2 is the charge of q2 (-2.0 μC), and r2 is the distance from q2 to q3 (0.50 m).

V2 = (9.0 x 10^9 N m^2/C^2) * (-2.0 x 10^-6 C) / 0.50 m.

V2 = -9.0 x 10^9 N m^2/C^2 * 0.004 C / 0.50 m.

V2 = -72 V.

To find the total electrical potential at the location of q3, we add the contributions from q1 and q2:

V_total = V1 + V2.

V_total = 180 V + (-72 V).

V_total = 108 V.

Therefore, the total electrical potential at the location of q3 created by q1 and q2 is 108 V.

b) The change in electric potential energy is given by the formula:

ΔPE = q3 * ΔV,

where ΔV is the change in electric potential, and q3 is the charge of q3 (+6.0 μC).

ΔPE = (6.0 x 10^-6 C) * (V_total - 0),

since the electric potential at infinity is usually taken to be zero.

ΔPE = (6.0 x 10^-6 C) * (108 V - 0).

ΔPE = 0.648 J.

Therefore, the change in electric potential energy in bringing charge q3 to this location is 0.648 J.

c) The work required to bring q3 to the location described is equal to the change in electric potential energy:

Work = ΔPE.

Work = 0.648 J.

Therefore, the work required to bring charge q3 to the location described is 0.648 J.

To determine the total electrical potential at the location of q3 created by charges q1 and q2, we can use the principle of the superposition of electric potentials. The electric potential at a point due to multiple charges is the algebraic sum of the electric potentials due to each individual charge. The electric potential at a point due to a single point charge q is given by:

V = k * q / r

Where V is the electric potential, k is the electrostatic constant (k = 9 x 10^9 Nm²/C²), q is the charge, and r is the distance between the charge and the point where we want to calculate the potential.

a) To find the total electrical potential at the location of q3 created by charges q1 and q2, we need to calculate the potentials individually and then sum them up.

For q1:
V1 = k * q1 / r1

For q2:
V2 = k * q2 / r2

Since the charges are equidistant from q3 in an equilateral triangle, r1 = r2 = 0.50 m. Plugging in the values:

V1 = (9 x 10^9 Nm²/C²) * (5.0 x 10^-6 C) / (0.50 m)
V1 = 9 x 10^9 Nm²/C * 10^-6 C / 0.50 m
V1 = 1.8 x 10^4 Nm/C

V2 = (9 x 10^9 Nm²/C²) * (-2.0 x 10^-6 C) / (0.50 m)
V2 = -3.6 x 10^4 Nm/C

Now, summing up the individual potentials:
Total potential at q3 = V1 + V2
Total potential at q3 = (1.8 x 10^4 Nm/C) + (-3.6 x 10^4 Nm/C)
Total potential at q3 = -1.8 x 10^4 Nm/C

b) The change in electric potential energy in bringing the charge q3 to this location can be calculated using the formula:

ΔPE = q3 * ΔV

Where ΔPE represents the change in potential energy, q3 is the charge, and ΔV is the change in electrical potential.

In this case, q3 = 6.0 x 10^-6 C and ΔV is the total potential at q3, which we calculated to be -1.8 x 10^4 Nm/C. Plugging in the values:

ΔPE = (6.0 x 10^-6 C) * (-1.8 x 10^4 Nm/C)
ΔPE = -10.8 x 10^-2 Nm
ΔPE = -1.08 J (Joules)

The negative sign indicates that the electric potential energy decreases as q3 is brought from infinity to the specified location.

c) The work required to bring the charge q3 to the location described is equal to the change in potential energy. So, the work required can be calculated as:

Work = ΔPE

Substituting the value of ΔPE we calculated in part b:

Work = -1.08 J (Joules)

Therefore, the work required to bring the charge q3 to the specified location is -1.08 Joules.

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