A point charge ( m = 1.0 g) at the end of an insulating string of length 51 cm i

Question

A point charge (m = 1.0 g) at the end of an insulating string of length 51 cm is observed to be in equilibrium in a uniform horizontal electric field of 9400 N/C, when the pendulum's position is as shown in Fig. 16-66, with the charge 1.0 cm above the lowest (vertical) position. If the field points to the right in Fig. 16-66, determine the magnitude and sign of the point charge.
_______Magnitude C
Sign

Negative, because the charge moves in the direction opposite to the electric field.

Positive, because the charge moves in the direction of the electric field.

Explanation / Answer

here,

mass , m = 1 g = 0.001 kg

length , l = 51 cm = 0.51 m

electric feild , E = 9400 N/C

theta = arccos((0.51 -0.01)/0.51)

theta = 11.4 degree

let the tension in the cable be T

equating the forces vertically

T * cos(theta) = m* g

T * cos(11.4) = 0.001 * 9.81

T = 0.01 N

equaitng the forces horizontally

T * sin(theta) = q * E

0.01 * sin(11.4) = q * 9400

q = 2.1 * 10^-7 C

the magnitude of charge is 2.1 * 10^-7 C

the sign is Positive, because the charge moves in the direction of the electric field.

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.