Question

A charged polystyrene ball of mass m = 10.0 g is hanging from the ceiling by a
piece of string with negligible mass. The ball hangs between two charged,
parallel plates which create a uniform electric field of magnitude E = 10.00 N/C
directed from left to right, as shown in the figure. When the ball is in
equilibrium between the plates, the string makes an angle of 20° to the vertical.
Calculate how many electrons have been added or removed from the
polystyrene ball to give it a nett charge. Specifically indicated if electrons have
been added or removed

Answer 1

Solution :

Here we have :

m = 10 g = 10 x 10^-3 kg

E = 10 N/C

θ = 20°

Let the charge on the charged polystyrene ball be Q.

And, Tension in the string be T.

.

.

From the given free-body diagram :

For equilibrium : T cosθ = F_g = mg

and, T sinθ = F_e

_.

From above equations :

Where, F_e is the electrostatic force on the charge.

And, F_e = Q E

Now, According to the Quantization of charge.

Q = n e

Thus, Number of electrons removed : n = Q / e = (3.57 x 10^-3 C) / (1.6 x 10^-19 C) = 2.23 x 10¹⁶ electrons

.

Since, The force due to electric field on the charge is along the direction of the electric field - The charge on the ball must be Positive. Which means - Electrons must be removed from the ball to get net positive charge.

And, Number of electrons that must be removed will be : 2.23 x 10¹⁶ electrons.