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An accelration produces a narrow beam of...

An accelration produces a narrow beam of protons, each having an initial speed of `v_(0)`. The beam is directed towards an initially uncharges distant metal sphere of radius R and centered at point O. The initial path of the beam is parallel to the axis of the sphere at a distance of `(R//2)` from the axis, as indicated in the diagram.

The protons in the beam that collide with the sphere will cause it to becomes charged. The subsequentpotential field at the accelerator due to the sphere can be neglected. The angular momentum of a particle is defined in a similar way to the moment of a force. It is defined as the moment of its linear momentum, linear replacing the force. We may assume the angular momentum of a proton about point O to be conserved. Assume the mass of the proton as `m_(P)` and the charge on it as e. Given that the potential of the sphere increases with time and eventually reaches a constant velue.
The total energy (E) of a proton in the beam travelling with seed v at a distance of `r (r ge R)` from point O. Assuming that the sphere has acquired an electrostatic charge Q is

A

`(eQ)/(4piepsilon_(0)r)`

B

less than `(eQ)/(4pi epsilon_(0)r)`

C

greater than `(eQ)/(4pi epsilon_(0)r)`

D

zero

Text Solution

Verified by Experts

The correct Answer is:
C
`P.E=(1)/(4pi in_(0))Q_(e)/r` and `KE=1/2 mv^(2)`
Hence total energy is greater than `(eQ)/(4pi in_(0)r)`
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