Q#23.23
(a) An electron is to be accelerated from 3.00 $\times 10^6$ m/s to 8.00 $\times 10^6$ m/s. Through what potential difference must the electron pass to accomplish this?
(b) Through what potential difference must the electron pass if it is to be slowed from 8.00 $\times 10^6$ m/s to a halt?
Answer:
Apply conservation of energy, Eq.
$K_1+U_1=K_2+U_2$
Use U = qV to express U in terms of V.
(a) $K_1+qV_1=K_2+qV_2$
$q(V_1-V_2)=K_1-K_2$ for q = $-1.602 \times 10^{-19} \ C$
$K_1= \frac{1}{2}m_ev_1^2=4.099 \times 10^{-18}$ J
$K_2= \frac{1}{2}m_ev_2^2=2.915 \times 10^{-17}$ J
$\Delta V = V_2-V_1=\frac{K_1-K_2}{q}$
$\Delta V = V_2-V_1=\frac{4.099 \times 10^{-18} \ J-2.915 \times 10^{-17} \ J}{-1.602 \times 10^{-19} \ C}$
$\Delta V=156$ V
The electron gains kinetic energy when it moves to higher potential.
Q#23.24
At a certain distance from a point charge, the potential and electric-field magnitude due to that charge are 4.98 V and 12.0 V/m respectively. (Take the potential to be zero at infinity.)
(a) What is the distance to the point charge?
(b) What is the magnitude of the charge?
(c) Is the electric field directed toward or away from the point charge?
Answer:
For a point charge, $E=k\frac{|q|}{r^2}$ and $E=k\frac{q}{r}$.
The electric field is directed toward a negative charge and away from a positive charge.
(a) V > 0 so q > 0
$\frac{V}{E}=\frac{kq/r}{k|q|/r^2}$
$\frac{V}{E}=\left(\frac{kq}{r}\right)\left(\frac{r^2}{kq}\right)=r$
$r=\frac{4.98 \ V}{12.0 \ V/m}$ = 0.415 m
(b) $q=\frac{rV}{k}=\frac{(0.415 \ m)(4.98 \ V)}{8.99 \times 10^9 \ N.m^2/C^2}$
q = $2.30 \times 10^{-10} \ C$
(c) q > 0, so the electric field is directed away from the charge.
The ratio of V to E due to a point charge increases as the distance r from the charge increases, because E falls off as $1/r^2$ and V falls off as $1/r$.
Q#23.25
A uniform electric field has magnitude E and is directed in the negative x-direction. The potential difference between point a (at x = 0.60 m) and point b (at x = 0.90 m) is 240 V.
(a) Which point, a or b is at the higher potential?
(b) Calculate the value of E.
(c) A negative point charge q = $-0.200 \mu C$ is moved from b to a. Calculate the work done on the point charge by the electric field.
Answer:
(a) The direction of $\vec{E}$ is always from high potential to low potential so point b is at higher potential.
(b) Apply $V_a-V_b=q'\int_{a}^{b}\vec{E}.d\vec{l}=\int_{a}^{b}E \ cos \phi \ dl$ to relate $V_b-V_a$ to E.
$V_a-V_b=E(x_b-x_a)$
$E=\frac{V_b-V_a}{x_b-x_a}$
$E=\frac{+240 \ V}{0.90 \ m - 0.60 \ m}$ = 800 V/m
(c) $W_{a→b}= q(V_b-V_a)$
$W_{a→b}= (-0.200 \times 10^{-6} \ C)(+240 \ V)=-4.80 \times 10^{-5}$ J
The electric force does negative work on a negative charge when the negative charge moves from high potential (point b) to low potential (point a).
Q#23.26
For each of the following arrangements of two point charges, find all the points along the line passing through both charges for which the electric potential V is zero (take V = 0 infinitely far from the charges) and for which the electric field E is zero:
(a) charges +Q and +2Q separated by a distance d and
(b) charges $-$Q and +2Q separated by a distance d.
(c) Are both V and E zero at the same places? Explain.
Answer:
For a point charge, $V=kq/r$. The total potential at any point is the algebraic sum of the potentials of the two charges. For a point charge, $E=k|q|/r^2$.
The net electric field is the vector sum of the electric fields of the two charges.
$\vec{E}$ produced by a point charge is directed away from the point charge if it is positive and toward the charge if it is negative.
(a) $V=V_Q+V_{2Q}>0$ so V is zero nowhere except for infinitely far from the charges. The fields can cancel only between the charges, because only there are the fields of the two charges in opposite directions. Consider a point a distance x from Q and d − x from 2Q, as shown in Figure 23.26a.
$E_Q=E_{2Q}$
$\frac{kQ^2}{x^2}=\frac{k(2Q)}{(d-x)^2}$
$(d-x)^2=x^2$
$x=\frac{d}{1+\sqrt{2}}$
The other root, $x=\frac{d}{1-\sqrt{2}}$, does not lie between the charges.
(b) V can be zero in 2 places, A and B, as shown in Figure 23.26b. Point A is a distance x from −Q and d − x from 2Q. B is a distance y from −Q and d + y from 2Q.
At A: $\frac{k(-Q)}{x}+\frac{k(2Q)}{d-x}=0$
$-\frac{1}{x}+\frac{2}{d-x}=0$
$x=\frac{d}{3}$
At B: $\frac{k(-Q)}{y}+\frac{k(2Q)}{d+y}=0$
$-\frac{1}{x}+\frac{2}{d-x}=0$
$y=d$
The two electric fields are in opposite directions to the left of −Q or to the right of 2Q in Figure 23.26c. But for the magnitudes to be equal, the point must be closer to the charge with smaller magnitude of charge. This can be the case only in the region to the left of −Q.
$E_Q=E_{2Q}$ gives
$\frac{kQ^2}{x^2}=\frac{k(2Q)}{(d+x)^2}$
$(d+x)^2=x^2$
$x=\frac{d}{\sqrt{2}-1}$
(d) E and V are not zero at the same places. $\vec{E}$ is a vector and V is a scalar. E is proportional to $1/r^2$ and V is proportional to 1/r. $\vec{E}$ is related to the force on a test charge and ΔV is related to the work done on a test charge when it moves from one point to another.
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