Answer:
Electric field at a distance of 1.45 cm will be
Explanation:
We have given the distance d = 1.45 cm = 0.0145 m
And the potential difference
There is a relation between potential difference and electric field
Electric field at a distance d due to a potential difference is given by
, here E is electric field, V is potential difference and d is distance
So
Explanation:
It is given that,
Frequency of vibration, f = 215 Hz
Amplitude, A = 0.832 mm
(a) Let T is the period of this motion. It is given by the following relation as :
(b) Speed of sound in air, v = 343 m/s
It can be given by :
Hence, this is the required solution.
Answer:
Velocity of the electron at the centre of the ring,
Explanation:
<u>Given:</u>
Total charge of the ring
Potential due the ring at a distance x from the centre of the rings is given by
The potential difference when the electron moves from x=0.2 m to the centre of the ring is given by
Let be the change in potential Energy given by
Change in Potential Energy of the electron will be equal to the change in kinetic Energy of the electron
So the electron will be moving with