Question

1.

Background pertinent to this problem is available in Interactive LearningWare 18.3. A uniform electric field exists everywhere in the x, y plane. This electric field has a magnitude of 3900 N/C and is directed in the positive x direction. A point charge -6.8 × 10-9 C is placed at the origin. Find the magnitude of the net electric field at (a) x = -0.19 m, (b) x = +0.19 m, and (c) y = +0.19 m.

(a) Number   Units

(b) Number   Units

(c) Number   Units

2.

In a vacuum, two particles have charges of q₁ and q₂, where q₁ = +3.8C. They are separated by a distance of 0.32 m, and particle 1 experiences an attractive force of 4.6 N. What is the value of q₂, with its sign?

number units

3.

Two charges are placed on the x axis. One of the charges (q₁ = +6.70C) is at x₁ = +3.00 cm and the other (q₂ = -28.9C) is at x₂ = +9.00 cm. Find the net electric field (magnitude and direction given as a plus or minus sign) at (a) x = 0 cm and (b) x = +6.00 cm.

(a)

Number

Units

(b)

Number

Units

please help

Answer 1

Electric field, E = 3900 N/C acting along +ve X axis

Point charge placed at the origin, q = - 6.8 x 10^-9 C

(a)

Electric field at the point x = -0.19 m due to charge q is E1 = 1/4πε₀ q/x²

E1 = 9 x 10⁹ x 6.8 x 10^-9 /0.19 x 0.19 = 1695.3 N/C

The net electric field at the given point is

E_net = E + E1 = 3900 + 1695.3 = 5595.3 N/C

The net electric field at the given point is E_net = 5595.3 N/C

(b)

Electric field at the point x = +0.19 m due to charge q is E2 = 1/4πε₀ q/x²

E1 = 9 x 10⁹ x 6.8 x 10^-9 /0.19 x 0.19 = 1695.3 N/C

The net electric field at the given point is

E_net = E + E2 = 3900 - 1695.3 = 2204.7 N/C

The net electric field at the given point is E_net = 2204.7 N/C

( c )

Electric field at the point xy= +0.19 m due to charge q is E3 = 1/4πε₀ q/x²

E1 = 9 x 10⁹ x 6.8 x 10^-9 /0.19 x 0.19 = 1695.3 N/C

The net electric field at the given point is

E_net =( E² + E3²)^1/2 = (3900² + 1695.3²)^1/2 = 4252.5 N/C

The net electric field at the given point is E_net = 4252.5 N/C