Question

A charge of 21.0 μC  is held fixed at the origin.

A) If a -5.50 μC charge with a mass of 3.60 g is released from rest at the position (0.925 m, 1.17 m), what is its speed when it is halfway to the origin?

B) Suppose the -5.50 μC charge is released from rest at the point x = 12(0.925m) and y = 12(1.17m). When it is halfway to the origin, is its speed greater than, less than, or equal to the speed found in part A?

C) Find the speed of the charge for the situation described in part B.

Answer 1

Distance between the charges is given as,   r = x² + y²

r = [(0.925 m)² + (1.17 m)²]

r = 2.224525 m²

r = 1.49 m

(A) What is its speed when it is halfway to the origin?

we know that, K.E = P.E

(1/2) m v² = k_e Q₁ Q₂ / r

where, m = mass of second charge = 3.60 x 10^-3 kg

Q₁ = charge on first particle = 21 x 10^-6 C

Q₂ = charge on second particle = 5.5 x 10^-6 C

then, we get

(0.5) (3.60 x 10^-3 kg) v² = [(9 x 10⁹ Nm²/C²) (21 x 10^-6 C) (5.5 x 10^-6 C)] / (1.49 m)

v² = (1.0395 Nm²) / [(1.8 x 10^-3 kg) (1.49 m)]

v = 387.5 m²/s²

v = 19.6 m/s

(B) Suppose the -5.50 C charge is released from rest at the point x = 12(0.925m) and y = 12(1.17m).

When it is halfway to the origin, then its speed is less than to the speed found in part A.

(C) Find the speed of the charge for the situation described in part B.

we know that, K.E = P.E

(1/2) m v² = k_e Q₁ Q₂ / r

where, r = distance between the charges = x² + y²

r = [(11.1 m)² + (14.04 m)²]

r = 320.3316 m²

r = 17.8 m

then, we get

(0.5) (3.60 x 10^-3 kg) v² = [(9 x 10⁹ Nm²/C²) (21 x 10^-6 C) (5.5 x 10^-6 C)] / (17.8 m)

v² = (1.0395 Nm²) / [(1.8 x 10^-3 kg) (17.8 m)]

v = 32.4 m²/s²

v = 5.7 m/s