Determine the potential difference between two charged parallel plates that are 0.20 cm apart and have an electric field strengt
2 answers:
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For these question, it has two separate equations: 2f(a) and f(2a) .
For f(2a) equations its x=2a, so you must substitute 2a into the f(x) equation
For 2f(a), it means the two time of f(a) equation with x=a, so you substitute a inti f(x) equation first, then you multiply it by 2.
Answer:
v = 5.75 x 10⁶ m/s
Explanation:
The radius (r) of the circular orbit taken by a charged particle is related to its speed perpendicular to a magnetic field of strength B, and is given by
r = --------------(i)
Where,
q = charge of the particle
m = mass of the particle
Making v subject of the formula in equation (i) above gives
v = -------------------(ii)
Given;
r = 20cm = 0.2m
B = 0.3T
v = unknown
q = charge of proton = 1.6 x 10⁻¹⁹ C
m = mass of the proton = 1.67 x 10⁻²⁷kg
Substitute the values of m, q, B and r into equation (ii) above to get;
v =
Solving for v gives:
v = 5.75 x 10⁶ m/s
Therefore, the velocity of the proton is 5.75 x 10⁶ m/s