Question

A small spherical insulator of mass 3.70 10-2 kg and charge +0.600 µC is hung by a thin wire of negligible mass. A charge of −0.900 µC is held 0.150 m away from the sphere and directly to the right of it, so the wire makes an angle θ with the vertical.

A) Find the angle.
B) Find the tension in the wire.

Answer 1

A) θ = 36.27°

B) T = 0.4327N

Step-by-step explanation:

A) If we imagine the free body diagram of this problem , and taking moments about the x-axis and equate to zero, we have;

ΣFx = F_el - T_x = 0

Now, From coulombs law, we know that F = k•q1•q2/r²

Thus, applying to this question, we have;

F_el - T_x gives

(k•q_bob•q_free)/d² - T_x = 0

Now, due to the angle θ, T_x when resolved, is T•sinθ

Thus, we have;

(k•q_bob•q_free)/d² - T•sinθ = 0

And so;

(k•q_bob•q_free)/d² = T•sinθ - - - (eq1)

Now, resolving about the y-axis and equating to zero gives;

ΣFy = T_y - m_bob•g = 0

When T_y is resolved, it gives T•cosθ

Thus;

T•cosθ - m_bob•g = 0

T•cosθ = m_bob•g

T = m_bob•g/cosθ - - - - (eq2)

Let's put m_bob•g/ cosθ for T in eq 1 to get;

(k•q_bob•q_free)/d² = (m_bob•g/ cosθ)•sinθ

This gives;

(k•q_bob•q_free)/d² =m_bob•g tanθ

Making tanθ the subject ;

tanθ = (k•q_bob•q_free)/(d².m_bob•g)

We are given;

q_bob = +0.600 µC = 0.6 x 10^(9)C

q_free = −0.900 µC = −0.9 x 10^(9)C (we will use positive in calculation as the negative sign shows that electric field points radially towards the charge.)

m_bob = 3.6 x 10^(-2) kg

d = 0.15m

k is coulombs constant with a standard value of 8.99 x 10^(9) N.m²/C²

Plugging in relevant values ;

tanθ = (8.99 x 10^(9)•0.6 x 10^(9)•0.9 x 10^(9))/(0.15²•3.6 x 10^(-2)•9.8)

tanθ = 0.7339

θ = tan^(-1)0.7339

θ = 36.27°

B) Let's put 36.27° for θ in eq 2.

Thus;

T = (3.6 x 10^(-2)•9.8)/(cos 36.27)

T = 0.4327N