Question

A wall has a negative charge distribution producing a uniformhorizontal electric field. A small plastic ball of mass .01kg carrying charge of -80 μC is suspended by an uncharged,nonconducting thread .30m long. The thread is attached to thewall and the ball hans in equilibrium in electric and gravitationalfields. Electric force on ball has magnitude of .032N
Calculate magnitude of electric field at ball's location dueto charged wall and show direction on x, y coordinate axes
Determine perpendicular distance from wall to center ofball
The string is cut
Calculate magnitude of resulting acceleration of ball and itsdirection
Describe resulting path of ball

Answer 1

Answer:

a)  E = -4 10² N / C , b) x = 0.093 m, c)     a = 10.31 m / s², θ=-71.9⁰

Explanation:

For that exercise we use Newton's second Law, in the attached we can see a free body diagram of the ball

X axis

             $F_{e}$ - $T_{x}$ = m a

Axis y

            $T_{y}$ - W = 0

Initially the system is in equilibrium, so zero acceleration

            Fe = $T_{x}$  

            T_{y} = W

Let us search with trigonometry the components of the tendency

            cos θ = T_{y} / T

            sin θ = $T_{x}$  / T

           T_{y} = cos θ

           $T_{x}$  = T sin θ

We replace

            q E = T sin θ

            mg = T cosθ

             

a) the electric force is

                $F_{e}$ = q E

                E = $F_{e}$ / q

                E = -0.032 / 80 10⁻⁶

                E = -4 10² N / C

b) the distance to this point can be found by dividing the two equations

                q E / mg = tan θ

                θ = tan⁻¹ qE / mg

Let's calculate

              θ = tan⁻¹ (80 10⁻⁶ 4 10² / 0.01 9.8)

              θ = tan⁻¹ 0.3265

               θ = 18 ⁰

               sin 18 = x/0.30

               x =0.30 sin 18

               x = 0.093 m

c) The rope is cut, two forces remain acting on the ball, on the x-axis the electric force and on the axis and the force gravitations

X axis

           $F_{e}$ = m aₓ

            aₓ = q E / m

           aₓ = 80 10⁻⁶ 4 10² / 0.01

           aₓ = 3.2 m / s²

Axis y

           W = m $a_{y}$

           a_{y} = g

           a_{y} = 9.8 m/s²

The total acceleration is can be found using Pythagoras' theorem

             a = √ aₓ² + a_{y}²

             a = √ 3.2² + 9.8²

             a = 10.31 m / s²

The Angle meet him with trigonometry

               tan θ = a_{y} / aₓ

               θ = tan⁻¹ a_{y} / aₓ

               θ = tan⁻¹ (-9.8) / 3.2

               θ = -71.9⁰

Movement is two-dimensional type with acceleration in both axes