- #1
JNOD
- 4
- 2
- Homework Statement
- A thin, light, spherical metal shell is suspended from a thread. The sphere hangs straight down when a charged rod is held at 3 cm to the left of the metal sphere. It can be concluded that:
A. the charge of the sphere is the opposite sign of the charge of the rod.
B. the charge of the sphere is the same sign as the charge of the rod.
C. the sphere is charged, but the sign of the charge is unknown.
D. the sphere is uncharged.
The answer is (b) according to answer key.
- Relevant Equations
- None
(A) incorrect, because opposite signs attract, and the sphere would've been drawn to the charged rod.
(B) correct, according to the answer key, but if the charge of the sphere and the charge of the rod are the same, then wouldn't they repel each other? I'm confused as to why this is the correct choice.
(C) incorrect, because if the sphere is charged, it will either move away or toward the charged rod, but it "hangs straight down" instead.
(D) incorrect, because even if the sphere is uncharged, the presence of a charged rod would induced a charge on the sphere.
Honestly I am baffled by this question. I'm not sure if there's something I'm not understanding. This is not conductive charging since no contact was implied. I'm assuming it has something to do with induction and polarization. Any hints or help is appreciated.
(B) correct, according to the answer key, but if the charge of the sphere and the charge of the rod are the same, then wouldn't they repel each other? I'm confused as to why this is the correct choice.
(C) incorrect, because if the sphere is charged, it will either move away or toward the charged rod, but it "hangs straight down" instead.
(D) incorrect, because even if the sphere is uncharged, the presence of a charged rod would induced a charge on the sphere.
Honestly I am baffled by this question. I'm not sure if there's something I'm not understanding. This is not conductive charging since no contact was implied. I'm assuming it has something to do with induction and polarization. Any hints or help is appreciated.