Themes > Science > Physics > Electromagnetism > Electrostatics > Electric field > Problems

Problem 1.1

Three point charges, q1 = - 4 nC, q2 = 5 nC, and q3 = 3 nC, are placed as in Fig. 1.4.
  
Figure 1.4: Three point charges
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If r1 = 0.5 m and r3 = 0.8 m, find the force on q2 due to the other two charges.

Solution:
We first find the force on q2 due to q1 :

F1 = k $\displaystyle{\frac{q_1q_2}{r_1^2}}$ = 9.0 x 109 x $\displaystyle{\frac{4\times 10^{-9}\cdot 5\times 10^{-9} }{(0.5){}^2}}$ = 7.2 x 10- 7 N ,

which is directed to the left. The force on q2 due to q3 is found as:

F3 = k $\displaystyle{\frac{q_3q_2}{r_3^2}}$ = 9.0 x 109 x $\displaystyle{\frac{3\times 10^{-9} \cdot 5\times 10^{-9} }{(0.8){}^2}}$ = 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.

Problem 1.2


The charges in Fig. 1.5, with q1 = q = - q2 , form what is called a dipole. Find the electric field a distance d along the x -axis.
  
Figure 1.5: An electric dipole
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Solution:
We resolve the contributions from q1 and q2 into components. For q1 ,

     (E1)x = k $\displaystyle{\frac{q_1}{r_1^2}}$cos $\displaystyle\theta$ = k$\displaystyle{\frac{q}{a^2+d^2}}$$\displaystyle{\frac{d}{\sqrt{a^2+d^2}}}$,   
     (E1)y = - k $\displaystyle{\frac{q_1}{r_1^2}}$sin $\displaystyle\theta$ = - k$\displaystyle{\frac{q}{a^2+d^2}}$$\displaystyle{\frac{a}{\sqrt{a^2+d^2}}}$,   

while for q2 ,
     (E2)x = - k $\displaystyle{\frac{q_2}{r_1^2}}$cos $\displaystyle\theta$ = - k$\displaystyle{\frac{q}{a^2+d^2}}$$\displaystyle{\frac{d}{\sqrt{a^2+d^2}}}$,   
     (E2)y = - k $\displaystyle{\frac{q_2}{r_1^2}}$sin $\displaystyle\theta$ = k$\displaystyle{\frac{q}{a^2+d^2}}$$\displaystyle{\frac{a}{\sqrt{a^2+d^2}}}$.   

The total electric field then has components
     Ex = (E1)x + (E2)x = 0,   
     Ey = (E1)y + (E2)y = - 2k$\displaystyle{\frac{qa}{\left(a^2+d^2\right){}^{3/2}}}$.   

Thus, the net electric field is directed in the - y direction and has a magnitude of - 2kqa/$\left(a^2+d^2\right)$3/2.


Problem 1.3
 

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 $\theta$ = 15 o with respect to the vertical, what is the net charge on the ball?

  
Figure 1.6: Charged ball in an electric field
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Solution:
We first examine in Fig. 1.7 the forces present on the ball.

  
Figure 1.7: Force diagram
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In equilibrium the net force on the ball is zero; we thus have, in the y direction,

Tcos $\displaystyle\theta$ - mg = 0 $\displaystyle\Rightarrow$ T = $\displaystyle{\frac{mg}{\cos\theta}}$,

and in the x direction,

qE - Tsin $\displaystyle\theta$ = 0 $\displaystyle\Rightarrow$ q = $\displaystyle{\frac{T\sin\theta}{E}}$ = $\displaystyle{\frac{mg\tan\theta}{E}}$.

With the numbers given, we find

q = $\displaystyle{\frac{0.002\cdot 9.8 \cdot\tan\,15^\circ}{1000}}$ = 5.2 x 10- 6 C .

Thus, the charge on the ball is 5.2 $\mu$ C.

Problem 1.4
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?

  
Figure 1.8: Three point charges
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Solution:
We will first calculate the electric potential V due to a point charge q a distance r away as

V = k $\displaystyle{\frac{q}{r}}$,

which assumes the potential vanishes at infinity. The potential due to q2 at the point occupied by q1 is thus

V2 = k $\displaystyle{\frac{q_2}{r_2}}$ = 9.0 x 109 x $\displaystyle{\frac{2\times 10^{-9} }{0.03}}$ = 600 V ,

while that due to q3 is

V3 = k $\displaystyle{\frac{q_3}{r_3}}$ = 9.0 x 109 x $\displaystyle{\frac{-4\times 10^{-9} }{0.05}}$ = - 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$\displaystyle\Delta$ V = - q1 x (Vf - Vi) = - 8 x 10- 9 x $\displaystyle\left[0-(-120)\right]=$ - 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.



 
Problem 1.5
A constant electric field E = 2000 V/m exists as in Fig. 1.9. A 10 $\mu$ C charge of mass 20 g, initially at rest at x = - 1 m, is released. What is its speed at x = 5 m?

  
Figure 1.9: Point charge in a constant electric field
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Solution:
We first find the potential difference between the two points of interest:

E = - $\displaystyle{\frac{\Delta\,V}{\Delta\,x}}$ = - $\displaystyle{\frac{V_f-V_i}{x_f-x_i}}$ $\displaystyle\Rightarrow$ $\displaystyle\Delta$ V = - E$\displaystyle\Delta$ x = - 2000$\displaystyle\left[5-(-1)\right]=$ - 12,000 V .

The work done by the electric field in moving the charge is

W = - q$\displaystyle\Delta$ V = - 10 x 10- 6 $\displaystyle\cdot$ (- 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 = $\displaystyle\Delta$ K = Kf - Ki = $\displaystyle{\textstyle\frac{1}{2}}$mv 2 - 0 $\displaystyle\Rightarrow$ v = $\displaystyle\sqrt{\frac{2W}{m}}$ = $\displaystyle\sqrt{\frac{2\cdot 0.12}{0.02}}$ = 3.46 m / s .

Thus, the final speed of the particle is 3.46 m/s.


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