Q#23.13
A small particle has charge $-5.00 \mu C$ and mass 2.00 $\times 10^{-4}$ kg. It moves from point where the electric potential is to point where the electric potential is $V_A=+200 \ V$. The electric force is $V_B=+800 \ V$ the only force acting on the particle. The particle has speed 5.00 m/s at point What is its speed at point Is it moving faster or slower at than at Explain.
Answer:
Apply conservation of energy to points A and B.
$K_A+U_A=K_B+U_B$
with $U = qV$, so
$K_A+qV_A=K_B+qV_B$
$K_B=K_A+q(V_A-V_B)$
$K_B=0.00250 \ J+(-5\times 10^{-6} \ C)(200 \ V-800 \ V)$
$K_B=0.00550$ J
$\frac{1}{2}mv_B^2=0.00550$ J
$v_B=\sqrt{\frac{2K_B}{m}}$
$v_B=\sqrt{\frac{2(0.00550 \ J)}{2.00 \times 10^{-4}\ kg}}=7.42$ m/s
It is faster at B; a negative charge gains speed when it moves to higher potential.
Q#23.14
A particle with a charge of +4.20 nC is in a uniform electric field $\vec{E}$ directed to the left. It is released from rest and moves to the left; after it has moved its kinetic energy is found to be +1.50 $\times 10^{-6} \ J$.
(a) What work was done by the electric force?
(b) What is the potential of the starting point with respect to the end point?
(c) What is the magnitude of $\vec{E}$?
Answer:
The work-energy theorem says
$W_{a→b}= K_b-K_a$ and $\frac{W_{a→b}}{q}=V_a-V_b$
Point a is the starting point and point b is the ending point. Since the field is uniform,
$W_{a→b}= Fs \ cos \phi=E|q|s \ cos \phi$
The field is to the left so the force on the positive charge is to the left. The particle moves to the left so $\phi=0^0$ and the work $W_{a→b}$ is positive.
(a) $W_{a→b}= K_b-K_a=1.50 \times 10^{-6} \ J-0=1.50 \times 10^{-6}$ J
(b) $\frac{W_{a→b}}{q}=V_a-V_b$
$\frac{1.50 \times 10^{-6} \ J}{4.20 \times 10^{-9} \ C}=V_a-V_b$
$V_a-V_b=$ 357 V. Point a is at higher potential than point b.
(c) $W_{a→b}=E|q|s \ cos 0^0$
$E=\frac{W_{a→b}}{|q|s}=\frac{V_a-V_b}{s}$
$E=\frac{357 \ V}{6.00 \times 10^{-2} \ m}$
$E = 5.95 \times 10^3$ V/m
A positive charge gains kinetic energy when it moves to lower potential; $V_b<V_a$
Q#23.15
A charge of 28.0 nC is placed in a uniform electric field that is 4.00 $\times 10^4$ V/m directed vertically upward and has a magnitude of What work is done by the electric force when the charge moves
(a) 0.450 m to the right;
(b) 0.670 m upward;
(c) 2.60 m at an angle $45.0^0$ of downward from the horizontal?
Answer:
Apply the equation that precedes Eq. (23.17):
$W_{a→b}=q'\int_{a}^{b}\vec{E}.\vec{dl}$
Use coordinates where + y is upward and +x is to the right. Then $\vec{E}=E\hat{j}$ with $E=4.00 \times 10^4$ N/C.
(a) The path is sketched in Figure 23.15a.
$\vec{E}.\vec{dl}=(E.\hat{j}).(dx.\hat{i})=0$
$W_{a→b}=q'\int_{a}^{b}\vec{E}.\vec{dl=0}$
The electric force on the positive charge is upward (in the direction of the electric field) and does no work for a horizontal displacement of the charge.
(b) The path is sketched in Figure 23.15b.
$\vec{E}.\vec{dl}=(E.\hat{j}).(dy.\hat{j})=E.dy$
$W_{a→b}=q'\int_{a}^{b}\vec{E}.\vec{dl=q'E\int_{a}^{b}dy=q'E(y_b-y_a)}$
$y_b-y_a$ = +0.670 m, positive since the displacement is upward and we have taken +y to be upward.
$W_{a→b}=q'E(y_b-y_a)$
$W_{a→b}=(+28.0 \times 10^{-9} \ C)(4.00 \times 10^4 \ N/C)(+0.670 \ m)$
$W_{a→b}=+7.50 \times 10^{-4}$ J
The electric force on the positive charge is upward so it does positive work for an upward displacement of the charge.
(c) The path is sketched in Figure 23.15c.
$y_a=0$ and $y_b=-r \ sin \theta=-(2.60 \ m) \ sin \ 45^0=-1.838 \ m$. The vertical component of the 2.60 m displacement is 1.838 m downward.
$d\vec{l}=dx \hat{i}+dy \hat{j}$. (The displacement has both horizontal and vertical components.)
$\vec{E}.d\vec{l}=(E.\hat{j}).(dx \hat{i}+dy\hat{j})=E.dy$
(Only the vertical component of the displacement contributes to the work.)
$W_{a→b}=q'\int_{a}^{b}\vec{E}.\vec{dl=q'E\int_{a}^{b}dy}=q'E(y_b-y_a)$
$W_{a→b}=q'E(y_b-y_a)$
$W_{a→b}=(+28.0 \times 10^{-9} \ C)(4.00 \times 10^4 \ N/C)(-1.838 \ m)$
$W_{a→b}=-2.06 \times 10^{-3}$ J
The electric force on the positive charge is upward so it does negative work for a displacement of the charge that has a downward component.
Q#23.16
Two stationary point charges +3.00 nC and +2.00 nC are separated by a distance of An electron is released from rest at a point midway between the two charges and moves along the line connecting the two charges. What is the speed of the electron when it is 10.0 cm from the +3.00 nC charge?
Answer:
Apply $K_a+U_a=K_b+U_b$
Let, $q_1=+3.00 \ nm$ and $q_2=+2.00 \ nm$. At point a, $r_{1a}=r_{2a}$ = 0.250 m.
At point b, $r_{1b}$=0.100 mand $r_{2b}$ = 0.400 m. The electron has q = $-e$ and $m_e=9.11 \times 10^{-31} \ kg$.
$K_a=0$ since the electron is released from rest.
$-\frac{keq_1}{r_{1a}}-\frac{keq_2}{r_{2a}}=-\frac{keq_1}{r_{1b}}-\frac{keq_2}{r_{2b}}+\frac{1}{2}m_ev_b^2$
$E_a=K_a+U_a$
$E_a=k(-1.60 \times 10^{-19} \ C)\left(\frac{3.00 \times 10^{-9} \ C}{0.250 \ m}+\frac{2.00\times 10^{-19} \ C}{0.250 \ m}\right)$
$E_a=-2.88 \times 10^{-17}$ J
$E_b=K_b+U_b$
$E_a=k(-1.60 \times 10^{-19} \ C)\left(\frac{3.00 \times 10^{-9} \ C}{0.100 \ m}+\frac{2.00\times 10^{-19} \ C}{0.400 \ m}\right)+\frac{1}{2}m_ev_b^2$
$E_a=-5.07 \times 10^{-17}\ J + \frac{1}{2}m_ev_b^2$
Setting $E_a=E_b$ gives
$E_a=-5.07 \times 10^{-17}\ J + \frac{1}{2}m_ev_b^2=-2.88\times 10^{-17}$ J
$v_b=\sqrt{\frac{2(5.04 \times 10^{-17} \ J-2.88 \times 10^{-17} \ J)}{9.11 \times 10^{-31} \ kg}}$
$v_b=6.89 \times 10^6$ m/s
$V_a=V_{1a}+V_{2a}=+180 \ V$ and $V_b=V_{1b}+V_{2b}=+315 \ V$
$V_b>V_a$. The negatively charged electron gains kinetic energy when it moves to higher potential.
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