Question

A small m=1.81E-3kg plastic ball is suspended by a l=20.0cm long string in a uniform electric field, as shown below. If the ball is in equilibrium when the string makes a θ=12.8deg angle with the vertical, what is the net charge on the ball? Take E=1.84E+3iN/C.

Answer 1

Answer Explanation:

Free body diagram:

Initially, we have drawn a free–body diagram of the forces acting on the ball are classified as follows:

1. The string tension T directed along the string;
2. The force of gravity, mg, downward;
3. The electric force which must be parallel to the electric and so here it must point towards the right.

These forces are shown in figure. The magnitude of the electric force is q E, where q is the charge on the plastic ball; this charge must be positive since the force points in the same direction as E.

The ball is in equilibrium so the (vector) sum of the forces is zero. The condition that the vertical force components sum to zero allows us to Find T :

T cos 12.8 − mg = 0  ⇒   T= mg / cos 15.0 = (1.81 x 10^-3)(9.80) / (cos 12.8) = 1.81× 10^-2N

The condition that the horizontal forces sum to zero gives us:

−T sin 12.8 + Felec = 0

=> −T sin 12.8 + q E = 0

⇒   q = T sin 12.8 / E

Substitute values in the above equation:
q = (1.81 × 10^-2 N) sin 12.8 / (1.84 × 10³ NC ) = 2.18 × 10^-6 C = 2.18 µC (Answer)

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