In: Physics
A. The electric potential at a position located a distance of 19.2 mm from a positive point charge of 7.80×10-9C and 13.0 mm from a second point charge is 1.14 kV. Calculate the value of the second charge.
B. The potential difference between two parallel conducting plates in vacuum is 590 V. An alpha particle with mass of 6.50×10-27 kg and charge of 3.20×10-19 C is released from rest near the positive plate. What is the kinetic energy of the alpha particle when it reaches the other plate? The distance between the plates is 45.0 cm.
SOLUTION:
lets the second charge is =
now, according to question:
THE 2nd charge is
2)
m=6.50*10-27 kg
W=work done by partial
initial kinetic energy
final kinetic energy at reaches destination position
we know that,
Kinetic energy at reches position is =