Question

Three charged particles are positioned in the xy plane: a 50-nC charge at y = 6 m on the y axis, a –80-nC charge at x = –4 m on the x axis, and a 70-nc charge at y = –6 m on the y axis. What is the electric potential (relative to a zero at infinity) at the point x = 8 m on the x axis? Group of answer choices +48 V +81 V –18 V –72 V +5.8 V

Answer 1

Answer:

The answer is + 4.8V

Explanation:

The electric potential is

         V = k ∑ $q_{i} / r_{i}$

We apply this expression to our case with three charges

        V = k (q₁ / r₁ + q₂ / r₂ + q₃ / r₃)

The distance is

        R = √((x-x₀)² + (y-y₀)²+ (z-z₀)²)

Let's find the distances

q₁ = 50 10⁻⁹ C

r₁ = √ ((8-0)² + (0-6)²)

r₁ = 10m

q₂ = -80 10⁻⁹ C

r₂ = √ ((8 + 4)²

r₂ = 12m

q₃ = 70 10⁻⁹ C

r₃ = √ ((8-0)²+ (0 + 6)²)

r₃ = 10 m

We calculate the potential

    V = 8.99 10⁸ 10⁻⁹ (50/10 -80/12 +70/10)

    V = 4.795 V

The answer is + 4.8V

Answer 2

Answer:

V = 48V

Explanation:

To get the electric potential (relative to a zero at infinity), we will need to sum the potentials due to all the point charges on the point in question.

Thus;

V = V1 + V2 + V3

Now, let's calculate the distance first.

d1 = (0,6) -> (8,0) =√[(8-0)² + (0-6)²] = √(64 + 36) = √100 = 10

d2 = (-4,0) -> (8,0) = √[(8-(-4))² + (0-0)²] ==√(12²) = 12

d3 = (0,-6) -> (8,0) = √[(8-0)² + (0-(-6))²] = √(64 + 36) = √100 = 10

Now to the potentials ;

V = kQ/r, where k is a constant with a value of 8.99 x 10^(9) N.m²/C²

From the question,

Q1 = 50nc = 50 x 10^(-9) C

Q2 = 80nc = 80 x 10^(-9)C

Q3 = 70nc = 70 x 10^(-9)C

Thus;

V1 = kQ1/d1 = (8.99 x 10^(9) x 50 x 10^(-9))/ 10m = 44.9V

V2 = kQ2/d2 = (8.99 x 10^(9) x 80 x 10^(-9))/ 12m = -59.9V

V3 = kQ3/d3 = (8.99 x 10^(9) x 70 x 10^(-9))/ 10m = 62.9V

Hence ;

V = V1 + V2 + V3

V = 44.9 - 59.9 + 62.9 = 48V