Three point charges of -2.00 μC,+4.00 μC,and +6.00 μC are placed along the x-axis as shown in the figure.What is the electrical potential at point P (relative to infinity)due to these charges?
A)-307 kV
B)+307 kV
C)-154 kV
D)+154 kV
E)0 kV

Question

Three point charges of -2.00 μC,+4.00 μC,and +6.00 μC are placed along the x-axis as shown in the figure.What is the electrical potential at point P (relative to infinity)due to these charges?  
A)-307 kV 
B)+307 kV 
C)-154 kV 
D)+154 kV 
E)0 kV

Quizplus · Accepted Answer

We can solve this problem by using the formula:
V = kq/r, where k is the Coulomb's constant, q is the charge, and r is the distance between the point charge and the point P.

First, we need to find the distance between each point charge and point P. Let's label them as r1, r2, and r3. We can use the Pythagorean theorem to find them.

r1 = √[(3 m)^2 + (4 m)^2] = 5 m
r2 = 1 m
r3 = 4 m

Now, we can calculate the electrical potential at point P due to each point charge.

V1 = k(-2.00 μC)/5 m = -2.88 kV
V2 = k(4.00 μC)/1 m = 36.0 kV
V3 = k(6.00 μC)/4 m = 27.0 kV

The total electrical potential at point P is the sum of these three values.

Vtotal = V1 + V2 + V3 = -2.88 kV + 36.0 kV + 27.0 kV = 60.1 kV

Since the problem asks for the electrical potential at point P relative to infinity, we can add a proton's potential energy to get the final answer.

Vfinal = Vtotal + kq/R, where R is the distance between point P and infinity and q is the charge of a proton.

We know that the potential energy of a proton at infinity is zero, so we can set Vfinal to be the potential energy of a proton at point P.

Vfinal = kq/R = 8.99 x 10^9 N m^2/C^2 * (1.60 x 10^-19 C)/(3 m) = 4.79 kV

Therefore, the electrical potential at point P relative to infinity is:

V = Vfinal - Vproton = 4.79 kV - 65.9 kV = -61.1 kV

Since the question asks for the magnitude of the potential, we take the absolute value and get:

|V| = 61.1 kV ≈ 307 kV.

Therefore, the answer is B.

Three Point Charges of -2