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Three point charges, q_{1} = - 4 nC, q_{2} = 5 nC, and
q_{3} = 3 nC, are placed as in Fig. 1.4.
If r_{1} = 0.5 m and r_{3} = 0.8 m, find the force on q_{2} due to the other two charges.
Solution:
We first find the force on q_{2} due to q_{1} :
F_{1} = k = 9.0 x 10^{9} x = 7.2 x 10^{- 7} N ,
which is directed to the left. The force on q_{2} due to q_{3} is found as:F_{3} = k = 9.0 x 10^{9} x = 2.11 x 10^{- 7} N ,
which is also directed to the left. Thus, the total force on q_{2} is given by7.2 x 10^{- 7} + 2.11 x 10^{- 7} = 9.31 x 10^{- 7} N .
Thus, the total force on q_{2} is 9.31 x 10^{- 7} N directed to the left.Solution:
We resolve the contributions from q_{1} and q_{2} into
components. For q_{1} ,
(E_{1})_{x} = k cos = k, | |||
(E_{1})_{y} = - k sin = - k, |
(E_{2})_{x} = - k cos = - k, | |||
(E_{2})_{y} = - k sin = k. |
E_{x} = (E_{1})_{x} + (E_{2})_{x} = 0, | |||
E_{y} = (E_{1})_{y} + (E_{2})_{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?
Solution:
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 findq = = 5.2 x 10^{- 6} C .
Thus, the charge on the ball is 5.2 C.
The three charges in Fig. 1.8, with q_{1} = 8 nC, q_{2} = 2 nC, and
q_{3} = - 4 nC, are separated by distances r_{2} = 3 cm and
r_{3} = 4 cm. How much work is required to move q_{1} to
infinity?
Solution:
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 q_{2} at the point occupied by q_{1} is thusV_{2} = k = 9.0 x 10^{9} x = 600 V ,
while that due to q_{3} isV_{3} = k = 9.0 x 10^{9} x = - 720 V ,
The net potential is thusV = 600 + (- 720) = - 120 V ,
and so the work required to move q_{1} to infinity isW = - q_{1} V = - q_{1} x (V_{f} - V_{i}) = - 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?
Solution:
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 isW = - 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 = K_{f} - K_{i} = mv^{ 2} - 0 v = = = 3.46 m / s .
Thus, the final speed of the particle is 3.46 m/s.www-admin@theory.uwinnipeg.ca