Question

In the figure, a small spherical insulator of mass 6.00×10−2 kg and charge +0.400 μC is hung by a thin wire of negligible mass. A charge of −0.220 μC is held 0.290 m away from the sphere and directly to the right of it, so the wire makes an angle theta with the vertical, as shown. What is the angle theta? (k=1/4πϵ0=8.99×109 N · m2/C2)

1.50∘

1.10∘

1.70∘

0.917∘

1.30∘

Answer 1

m = mass = 6*10^-2kg

q1 = charge on ist sphere = + 0.4*10^-6C

q2 = chaerge on second = - 0.22*10^-6C

r = distance = 0.290 m

let T = tension in wire

The electrostatic force is given by

$F = \frac{kq_1q_2}{r^2}$

@ = angle

Using the equillibrium condition, we get that

$T *\cos\theta = mg\\ T*\sin\theta= F$

Using the above two equations, we get that

$\tan\theta = F/ mg = \frac{kq_1q_2}{ mg*r^2}\\ = \frac{9*10^9*0.4*0.22*10^{-12}}{0.06*9.8*0.290^2}= 0.0160\\ \\ \theta = 0.917^\circ$

Ans: 0.917 deg