A point charge ( m = 1.0 g) at the end of an insulating string of length L = 46

Question

A point charge (m = 1.0 g) at the end of an insulating string of length L = 46 cm (Fig. 16-66) is observed to be in equilibrium in a uniform horizontal electric field of E = 9600 N/C, when the pendulum's position is as shown in Fig. 16-66, with the charge d = 2.0 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

A point charge (m - 1.0 g) at the end of an insulating string of length L 46 cm (Fig. 16-66) is observed to be in equilibrium in a uniform horizontal electric field of E 9600 N/C, when the pendulum's position is as shown in Fig. 16-66, with the charge d- 2.0 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 Sign O Negative, because the charge moves in the direction opposite to the electric field O Positive, because the charge moves in the direction of the electric field Figure 16-66

Explanation / Answer

The weight of the point charge is balanced by the vertical component of the tension

So, if the tension in the string is T, then its vertical component is TCos = T * ( adj / hyp ) (see the right triangle formed with the string as hypotenuse and vertical dotted line as a side)

Vertical component of T = T * ( (46 - 2) / 46 ) T * 44 /46

This should be equal to the weight of the point charge = 0.001 kg x 9.81 m /s2 = 0.00981 N

T* 44 / 46 = 0.00981

T = 0.010255909 N

Now the Horizontal component of T is balancing the Electrostatic force on the charge (for equilibrium)

Therefore, T sin = EQ

T Sin ( Cos -1 (44/46) ) = 9600 Q

T x 0.29166104 = 9600Q

0.010255909 x 0.29166104 = 9600Q

Q = 3.115884469 x 10-7 C

therefore, Q = 0.311588 C

Due to the gravity pendulum tries to fall down and the tension has a horizontal component acting towards the left, hence to balance that, the electrostatic force on the charge need to act towards right. which is possible only if the charge is positive, since the Electric field is towards right also.

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