Problems

$\fbox{\bf PROBLEM \thechapter.\arabic{probcount}}$

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

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

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

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





$\fbox{\bf PROBLEM \thechapter.\arabic{probcount}}$

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

Solution:
We resolve the contributions from q₁ and q₂ into components. For q₁ ,

$${\frac{q_1}{r_1^2}} \theta {\frac{q}{a^2+d^2}} {\frac{d}{\sqrt{a^2+d^2}}}$$

$${\frac{q_1}{r_1^2}} \theta {\frac{q}{a^2+d^2}} {\frac{a}{\sqrt{a^2+d^2}}}$$

while for q₂ ,

$${\frac{q_2}{r_1^2}} \theta {\frac{q}{a^2+d^2}} {\frac{d}{\sqrt{a^2+d^2}}}$$

$${\frac{q_2}{r_1^2}} \theta {\frac{q}{a^2+d^2}} {\frac{a}{\sqrt{a^2+d^2}}}$$

The total electric field then has components

$${\frac{qa}{\left(a^2+d^2\right){}^{3/2}}}$$

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.

$\fbox{\bf PROBLEM \thechapter.\arabic{probcount}}$

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

Solution:
We first examine in Fig. 1.7 the forces present on the ball.

  

Figure 1.7: Force diagram

In equilibrium the net force on the ball is zero; we thus have, in the y direction,

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

and in the x direction,

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

With the numbers given, we find

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

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

$\fbox{\bf PROBLEM \thechapter.\arabic{probcount}}$

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

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

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

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

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

while that due to q₃ is

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

$\fbox{\bf PROBLEM \thechapter.\arabic{probcount}}$

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

Solution:
We first find the potential difference between the two points of interest:

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

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

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

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