a = 2.8 × 10^8 m/s^2 , right.12 . F = 4.8 × 10−19 N, right.
a = 5.3 × 10^11 m/s^2 , right.13 . F = 4.8 × 10−19 N, left.
a = 5.3 × 10^11 m/s^2 , left.
Velocity does not affect electric force.14 . F = 4.8 × 10−19 N, left.
a = 5.3 × 10^11 m/s^2 , left.15 . F = 4.8 × 10−19 N, right.
a = 2.8 × 10^8 m/s^2 , right.16 . F = 4.8 × 10−19 N, left.
a = 5.3 × 10^11 m/s^2 , left.17 . F = 4.8 × 10−19 N, right.
a = 5.3 × 10^11 m/s^2 , right.Step-by-Step Solution to #2:
(a) The magnetic force on a charge is given by F = qvB sin θ . Since the velocity is perpendicular to the
magnetic field, θ = 90°, and sin θ = 1. The charge q is the amount of charge on an electron, 1.6 × 10−19 C.
v is the electron’s speed, 2 m/s. B is the magnetic field, 3 T.
F = (1.6 × 10−19 C)(2 m/s)(3 T)(1) = 9.6 × 10−19 NThe direction is given by the right-hand rule. Point your fingers in the direction of the electron’s velocity,
toward the top of the page; curl your fingers in the direction of the magnetic field, to the right; your thumb
points into the page. Since the electron has a negative charge, the force points opposite your thumb, or out
of the page.
(b) Even though we’re dealing with a magnetic force, we can still use Newton’s second law. Since the
magnetic force is the only force acting, just set this force equal to ma and solve. The direction of the
acceleration must be in the same direction as the net force.
9.6 × 10−19 N = (9.1 × 10−31 kg)a
a = 1.1 × 10^12 m/s^2 , out of the pageStep-by-Step Solution to #10:
(a) The electric force on a charge is given by F = qE . The charge q is the amount of charge on an
electron, 1.6 × 10−19 C. E is the electric field, 3 N/C.