Problem 16.16 11 of 11 tants Part A A large electroscope is made with \"leaves\"
ID: 1660513 • Letter: P
Question
Problem 16.16 11 of 11 tants Part A A large electroscope is made with "leaves" that are 78-cm-long wires with tiny 23 g spheres at the ends When charged, nearly all the charge resides on the spheres If the wires each make a 26 angle with the vertical (Figure 1), what total charge Q must have been applied to the electroscope? Ignore the mass of the wires Express your answer to two significant figures and include the appropriate units Figure 1 of1 Submit Previous Answers Request Answer X Incorrect; One attempt remaining: Try Again 6° 126 78 cm 78 cm Return to Assignment Provide FeedbackExplanation / Answer
There are three forces actin: Gravity acting downwards (Fg), electrostatic acting horizontally (Fe), and tension acting along the wire (T). The three must add up to 0 when you take the vector sum.
The only force for which we can make a direct calculation is Fg: Fg = mg. So we must ultimately find Fe in terms of Fg, then use Coulomb's Law to find the charge.
The electromotive force (Fe) must be equal and opposite the horizontal compontent of the tension (Tx): Fe = Tx. Find Tx in terms of T: T sin 26°. Therefore:
Fe = T sin 26°.
The gravitational force (Fg) must be equal and opposite the vertical component of tension (Ty): Fg = Ty. Find Ty in terms of T: T cos 26°. So:
Fg = T cos 26°.
Solve both for T:
Fe = T sin 26°
T = Fe / sin 26°
Fg = T cos 26°
T = Fg / cos 26°
Then set them equal and solve for Fe:
Fe / sin 26° = Fg / cos 26°
Fe = Fg sin 26° / cos 26°
Fe = Fg tan 26°
Fe = mg tan 26°
By Coulomb's Law, Fe is also given by: Fe = KQq / r², where K is the electrostatic constant, Q and q are the two charges, and r is the distance between the charges. This gives:
KQq / r² = mg tan 26°
Both charges are equal, so:
Kq² / r² = mg tan 26°.
Solve for q:
q² = mgr² tan 26° / K
q = (mgr² tan 26° / K)
Find r. The mass is deflected by a horizontal distance of: (78 cm) sin 30° = 0.78 sin 26°, but since both masses are deflected an equal amount, the distance between them is given by: r = (2)(0.78 sin 26°) = 0.68 m.
Finally, you can plug in your knowns (in SI units) and calculate:
q = [(0.023 kg)(9.8 m/s²)(0.68 m)²( tan 26°) / (8.988x10^9 Nm²/C²)
q ±2.39 x 10^-6 C = ±2.39 C
This is the charge on ONE sphere, so the total charge will be double this number.
2q ±4.8 C = 4.8 x 10-6 C(rounded to two significant digits)