The electric potential at point P is -420.12 V. The speed of the proton as it passes through the origin is 2.83 x 10^4 m/s.
Step-by-step explanation:
(a) Find the electric potential at point P:
The electric potential at a point due to two charges can be calculated using the equation:
V = k(q1/r1 + q2/r2)
Where k is the Coulomb's constant (8.99 x 10^9 Nm2/C2), q1 and q2 are the charges (-18.6 nC), and r1 and r2 are the distances between the charges and point P.
In this case, r1 = sqrt((-8.50)^2 + (0 - 2.50)^2) = 8.50 cm, and r2 = sqrt((-8.50)^2 + (0 - (-2.50))^2) = 8.50 cm.
Substituting these values into the equation, we get:
V = 8.99 x 10^9 (-18.6 x 10^-9 / 8.50 x 10^-2 + (-18.6 x 10^-9 / 8.50 x 10^-2) = -420.12 V
Therefore, the electric potential at point P is -420.12 V.
(b) Calculate the speed of the proton as it passes through the origin:
The electric potential energy of the proton at point P can be calculated using the equation:
U = qV
Where q is the charge of the proton (1.60 x 10^-19 C), and V is the electric potential at point P (-420.12 V).
Substituting these values into the equation, we get:
U = (1.60 x 10^-19 C) (-420.12 V) = -6.72 x 10^-17 J
The kinetic energy of the proton can be calculated using the equation:
K = (1/2)mv^2
Where m is the mass of the proton (1.67 x 10^-27 kg), and v is the speed of the proton at the origin.
Equating the potential energy to the kinetic energy, we get:
U = K
-6.72 x 10^-17 J = (1/2) (1.67 x 10^-27 kg) v^2
Simplifying the equation, we get:
v^2 = (-6.72 x 10^-17 J) / ((1/2) (1.67 x 10^-27 kg))
v^2 = -8.03 x 10^9 m2/s2
Taking the square root of both sides of the equation, we get:
v = 2.83 x 10^4 m/s
Therefore, the speed of the proton as it passes through the origin is 2.83 x 10^4 m/s.