Question

Two small plastic balls are given equal charges by rubbing them on rabbit fur. When the balls are moved 8.00 meters apart, they have a measured electrostatic force of +7.24 x 10-3 N between them. a. What is the charge on each ball? b. How many electrons did each ball take from the fur? c. What is the electric field at a point in space (Point P) exactly 5.00 meters from both balls. d. What is the electric PE of an electron placed at Point P? e. What is the acceleration of the electron if it is released from that point?

Answer 1

Answer:

Given, F = 7.24 x 10^-3 N and separation distance r = 8 m

(a) Electrostatic force F = k q²/r² which implies that q² = F r²/k = (7.24 x 10^-3 N) (8 m)² / (9 x 10⁹ N m²/C²)

Therefore, q² = 51.48 x 10^-12 C²    (or) q = 7.17 x 10^-6 C.

(b) Number of electrons calculated from q = ne (or) n = q/e = (7.17 x 10^-6 C) / (1.6 x 10^-19 C) = 4.48 x 10¹³.

(c) The elctric field due to the given two charge system is E_net = 2 E_y = 2 E cos 45⁰, where E_y is the verticle component and the horizontal componets E_x's due to two charges were cancelled.

Therefore, E_net = 2 (kq/r²) cos 45⁰ since E = kq/r² = (9 x 10⁹ N m²/C²) (7.17 x 10^-6 C) / (5 m)² = 2.58 x 10³ N/C.

Then, E_net = 2 (2.58 x 10³ N/C) cos 45⁰ = 3.64 x 10³ N/C.

(d) Electrostatic PE is U = kq²/r. The U is due to two charges at the point p is U = U_q + U_q = 2 U_q

Therefore, U = 2 (kq²/r) = 2 (9 x 10⁹ N m²/C²) (7.17 x 10^-6 C)² / (5 m) = 185.07 x 10^-3 J.

(e) If an electron is released from the point P, then the electric field E_net applies an elctric force on the elctron and then it begins to accelerate, therefore, F = qE = ma or a = qE/m = (7.17 x 10^-6 C) (3.64 x 10³ N/C)/(9.1 x 10^-31 kg)

Acceleration of the electron is a = 2.86 x 10²⁸ m/s².