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Three point charges, q1 = - 4 nC, q2 = 5 nC, and q3 = 3 nC, are placed as in Fig. 1.4.
If r1 = 0.5 m and r3 = 0.8 m, find the force on q2 due to the other two charges.
We first find the force on q2 due to q1 :
F1 = k = 9.0 x 109 x = 7.2 x 10- 7 N ,which is directed to the left. The force on q2 due to q3 is found as:
F3 = k = 9.0 x 109 x = 2.11 x 10- 7 N ,which is also directed to the left. Thus, the total force on q2 is given by
7.2 x 10- 7 + 2.11 x 10- 7 = 9.31 x 10- 7 N .Thus, the total force on q2 is 9.31 x 10- 7 N directed to the left.
We resolve the contributions from q1 and q2 into components. For q1 ,
|(E1)x = k cos = k,|
|(E1)y = - k sin = - k,|
|(E2)x = - k cos = - k,|
|(E2)y = - k sin = k.|
|Ex = (E1)x + (E2)x = 0,|
|Ey = (E1)y + (E2)y = - 2k.|
A m = 2 g ball is suspended by a l = 20 cm long string as in Fig. 1.6 in a constant electric field of E = 1000 N/C. If the string makes an angle of = 15 o with respect to the vertical, what is the net charge on the ball?
We first examine in Fig. 1.7 the forces present on the ball.
Tcos - mg = 0 T = ,and in the x direction,
qE - Tsin = 0 q = = .With the numbers given, we find
q = = 5.2 x 10- 6 C .Thus, the charge on the ball is 5.2 C.
The three charges in Fig. 1.8, with q1 = 8 nC, q2 = 2 nC, and q3 = - 4 nC, are separated by distances r2 = 3 cm and r3 = 4 cm. How much work is required to move q1 to infinity?
We will first calculate the electric potential V due to a point charge q a distance r away as
V = k ,which assumes the potential vanishes at infinity. The potential due to q2 at the point occupied by q1 is thus
V2 = k = 9.0 x 109 x = 600 V ,while that due to q3 is
V3 = k = 9.0 x 109 x = - 720 V ,The net potential is thus
V = 600 + (- 720) = - 120 V ,and so the work required to move q1 to infinity is
W = - q1 V = - q1 x (Vf - Vi) = - 8 x 10- 9 x - 9.6 x 10- 7 J .Thus, the work required is 9.6 x 10- 7 J, with the minus sign indicating that a net work must be supplied.
A constant electric field E = 2000 V/m exists as in Fig. 1.9. A 10 C charge of mass 20 g, initially at rest at x = - 1 m, is released. What is its speed at x = 5 m?
We first find the potential difference between the two points of interest:
E = - = - V = - E x = - 2000 - 12,000 V .The work done by the electric field in moving the charge is
W = - q V = - 10 x 10- 6 (- 12,000) = 0.12 J .The positive sign indicates the electric field is doing the work. This work goes into changing the kinetic energy of the particle:
W = K = Kf - Ki = mv 2 - 0 v = = = 3.46 m / s .Thus, the final speed of the particle is 3.46 m/s.