Ametal sphere hangs from a string and has 4 kg of mass. the sphere has a charge of +6.5 µc. a uniform electric field is turned on and directed to the right.

(a) which way does the sphere move?

(b) what is the electric force on the metal sphere if the magnitude of the electric field is 106 n/c?

(c) when the sphere reaches equilibrium, what is the angle that the string makes with the vertical axis (assuming the same electric field as part (

a) to the right. b) 6.5 N c) 9.4º

Explanation:

a) As the sphere is positively charged experiments a force that has the same direction of the applied electric field, which is the one that would take a positive test charge, so the sphere under the influence of this electric field, moves to the right.

b) By definition, the magnitude of the electric field id the force per unit charge, as follows:

E= F/Q

Solving for F, we find the electric force on the metal sphere, based on the information provided:

F = Q*E = 6.5*10⁻⁶ C*10⁶ N/C = 6.5 N

c) When the sphere reaches equilibrium, if we apply Newton´s 2nd Law to the forces acting on the sphere, we have.

Fext = 0

As this is a vector equation, we can decompose the force vector in 2 components mutually perpendicular to turn the vector equation in two algebraic expressions.

In this case, it is advisable to choose a pair of axes coincident with the horizontal direction (x-axis, we choose positive x going to the right), and vertical direction (y-axis, we choose positive y going up).

We have three external forces acting on the sphere: one horizontal (electric force, going to the right), one vertical (gravity, always downward) and the tension T on the string, that will have projections on both axes.

The projection on the vertical, will be the product of T times the cosine of the angle that the string forms with the vertical.

So, the equation for the y-axis is as follows:

Fy = T*cosθ -m*g = 0

For the horizontal, the projection of T on this axis is the product of T times the sine of the same angle, so we can write the following equation for the x-axis:

Fx = -T*sinθ +Q*E = 0

Dividing both sides of (1) and (2), we get:

tg θ = Q*E / m*g = 0.165

⇒ θ = 9.4º

54 degrees incident angle = reflected angle

angle is from normal turned to line of ray

find the sum of the two horizontal vectors and the sum of the two vertical vectors

isoceles right angled triangle