Question

A large positively charged object with charge q + = 3.25 μC q+=3.25 μC is brought near a negatively charged plastic ball suspended from a string of negligible mass. The suspended ball has a charge of q − = − 48.3 nC q−=−48.3 nC and a mass of 15.5 g. 15.5 g. What is the angle the string makes with the vertical when the positively charged object is 17.5 cm 17.5 cm from the suspended ball? The positively charged object is at the same height as the suspended ball.

Answer 1

Answer:

θ = 16.9º

Explanation:

In the attached drawing, we can see that the plastic ball is in equilibrium, under the influence of three external forces: gravity (downward), the attractive electrostatic force (directed horizontally to the right), and the tension in the string.

According Newton's 2nd Law, the sum of the external forces must add to 0, as the ball is in static equilibrium.

If we convert this vector equation, in two algebraic equations, finding the components of the forces along two perpendicular directions, both equations must add to 0 also.

Just for simplicity, we can choose these directions to be coincident with the horizontal (x-axis) and vertical (y-axis) directions.

The only force that have components on both axis, is the tension in the string, so, we need to find the projections of T over both axis, as follows:

Ty = T*cosθ

Tx = T*sin θ

Replacing these values in the equations for Fx and Fy, we have:

Fx = T* sinθ + $\frac{k*q1*(-q2)}{r^{2}}$ = 0  ⇒ T* sinθ = $\frac{k*q1*q2}{r^{2}}$ (1)

Fy = T * cos θ - m*g = 0 ⇒        T* cosθ = m*g (2)

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

tg θ =$\frac{k*q1*q2*r^{2}}{m*g} = \frac{9*10(9) N*m2/C2*3.25 uC*48.3nC}{0.0155 kg*9.8m/s2} = 0.3038$

⇒ θ = tg⁻¹ (0.3038) = 16.9º, being θ the angle showed in the attached picture.