Question

1) Calculate the values of E and the potential (V) at the third point for q1 = +10.0 µC and q2 = -10.0 µC and the distance between q1 and q2 = 10.0 cm.

2) Imagine there is a -2e-6 C charge on the third point. Caluclate the force on this charge. What is the direction of this force?

3) What is the path the charge might take if it is released? Assume initial velocity is zero.

Answer 1

1.

Along the X-direction

E_x = E₁ Cos60 + E₂ Cos60 = (kq₁/r² + kq₂/r²) Cos60 = (9 x 10⁹) (q₁ + q₂) Cos60/r²

E_x = (9 x 10⁹) (10 + 10) (10^-6) Cos60/(0.1)² = 9 x 10⁶ N/C

Along the Y-direction

E_y = E₁ Sin60 - E₂ Sin60 = (kq₁/r² - kq₂/r²) Sin60 = 0

Net electric field = E = E_x = 9 x 10⁶ N/C

Electric Potential is given as

V = V₁ + V₂ = k q₁/r + k q₂/r = (k/r) (q₁ + q₂) = (k/r) (10 - 10) = 0 volts

2)

F = qE = (2 x 10^-6) (9 x 10⁶) = 18 N

direction : since electric field is in Positive x-direction and charge "q" is negative hence Force experienced is in negative X-direction