Question

Chapter 18, Problem 21 Interactive LearningWare 18.2 provides one approach to solving problems such as this one. The drawing shows three point charges fixed in place. The charge at the coordinate origin has a value of q1 = +6.051C; the other two have identical magnitudes, but opposite signs: q2 = -4.8ß1C and q3 = +4.80 . (a) Determine the net force exerted on q1 by the other two charges. (b) Ifq1 had a mass of 1.50 g and it were free to move, what would be its acceleration? 92 1.30 m 23.0 3.0 1.30 m 43 (a) Number Units (b) Number Units

Answer 1

Givens:

$\large q_1=6.05\mu C=6.05\times10^-^6 C$

$\large q_2=4.8 \mu C=4.8 \times10^-^6 C$

$\large q_3=-4.8 \mu C=-4.8 \times10^-^6 C$

$\large r=1.3\,m$

$\large \theta=23^o$

mass, $\large m=1.5\,g=1.5\times10^-^3\,kg$

(a) There are two electrostatic forces that act on q1; that due to q2 and that due to q3. The magnitudes of these forces can be found by using Coulomb's law. The magnitude and direction of the net force that acts on q1 can be determined by using the method of vector components.

The magnitude $\large F_1_2$ of the force exerted on q1 by q2 is given by Coulomb's law, where the distance is specified in the drawing:

$\large F_1_2=\frac{kq_1q_2}{r_1_2^2}=\frac{8.99\times10^9\times6.05\times10^-^6\times4.8\times10^-^6}{1.3^2}=0.1545\,\,N$Since the magnitudes of the charges and the distances are the same, the magnitude of F13 is the same as the magnitude of F12,ie, F13 = 0.1545 N. From the drawing, it can be seen that the x-components of the two forces cancel, so we need only to calculate the y components of the forces. Moreover, since both the y-components point towards positive y axis, the final value will be twice this.

The y-component of the forces can be calculated by,

$\large F_y=F\,\,sin\,\,\theta$

Here, $\large ____F_1_2__y=F_1_3__y\,\, ^=0.1545\,\,sin\,\,23=0.6040$

  $\large F_n_e_t=2\times F_y=2\times0.0604=0.1208\,\,N$

(b) According to Newton's second law, the acceleration of q1 is equal to the net force divided by its mass.

  $\large a=\frac{F}{m}=\frac{0.1208}{1.5\times10^-^3}=80.5\,\,m/s$