Question

6)

Two charges of q_1 = 1.2 muC and q_2 = -2.0 are d = 0.66 m apart at two vertices of an equilateral triangle as in the figure below. (a) What is the electric potential due to the 1.2-muC charge at the third vertex, point P? v (b) What is the electric potential due to the 2.0-muC charge at P? V (c) Find the total electric potential at P. V (d) What is the work required to move a 3.5-muC charge from infinity to P? J

Answer 1

(a) The electric potential due to the 1.2 _$\mu$ C charge at the third vertex, point P will be given as :

V₁ = k_e q₁ / d = (9 x 10⁹ Nm²/C²) (1.2 x 10^-6 C) / (0.66 m)

V₁ = 1.63 x 10⁴ V

(b) The electric potential due to the -2 _$\mu$ C charge at the third vertex, point P will be given as :

V₂ = k_e q₂ / d = (9 x 10⁹ Nm²/C²) (2 x 10^-6 C) / (0.66 m)

V₂ = -2.72 x 10⁴ V

(c) The total electric potential at P which will be given as :

$\Delta$ V = V₁ + V₂ = [(1.63 x 10⁴ V) - (2.72 x 10⁴ V)]

$\Delta$ V = - 1.09 x 10⁴ V

(d) The work required to move a 3.5 _$\mu$ C charge from an infinity to P which will be given as :

using a formula, we have

$\Delta$ W = q $\Delta$ V = (3.5 x 10^-6 C) (-1.09 x 10⁴ V)

$\Delta$ W = - 3.81 x 10^-2 J