Question

A charge of +6.10 mC is located at x = 0, y = 0 and a charge of −8.80 mC is located at x = 0, y = 3.00 m. What is the electric potential due to these charges at a point P with coordinates x = 4.00 m, y = 0?

Answer 1

The electric potential due to these charges at a point P is 2.1 x 10⁷ Nm/C.

Distance between point P and charge at y = 3 m

Apply Pythagoras theorem to determine the distance between these two points.

y² = 3² + 4²

y² = 25

y = √25

y = 5 m

Electric potential at point P due to -8.8 mC

$V_y = \frac{kQ}{r} \\\\V_y = \frac{9\times 10^9 \times (-8.8 \times 10^{-3})}{5} \\\\V_y = -1.584 \times 10^{7} \ Nm/C$

Electric potential at point P due to +6.10 mC

$V_x = \frac{kQ}{r} \\\\V_x = \frac{9\times 10^9 \times 6.1 \times 10^{-3}}{4} \\\\V_x = 1.373 \times 10^{7} \ Nm/C$

Resultant electric potential

$V = \sqrt{V_y^2 + V_x^2} \\\\V = \sqrt{(-1.584 \times 10^7 )^2\ + \ (1.373 \times 10^7)^2} \\\\V = 2.1 \times 10^7 \ Nm/C$

Thus, the electric potential due to these charges at a point P is 2.1 x 10⁷ Nm/C.

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