Q#23.17
Point charges $q_1=2.00 \mu C$ and $q_2=-2.00 \mu C$ are placed at adjacent corners of a square for which the length of each side is 3.00 cm. Point a is at the center of the square, and point b is at the empty corner closest to $q_2$. Take the electric potential to be zero at a distance far from both charges.
(a) What is the electric potential at point a due to $q_1$and $q_2$?
(b) What is the electric potential at point b?
(c) A point charge $q_3=-5.00 \mu C$ moves from point a to point b. How much work is done on $q_3$ by the electric forces exerted by $q_1$ and $q_2$? Is this work positive or negative?
Answer:
The potential at any point is the scalar sum of the potentials due to individual charges.
$V=\frac{kq}{r}$ and $W_{ab}=q(V_a-V_b)$
(a) $r_{a1}=r_{a2}=\frac{1}{2}\sqrt{(0.0300 \ m)^2+(0.0300 \ m)^2}=0.0212 \ m$
$V_a=k\left(\frac{q_1}{r_{a1}}+\frac{q_2}{r_{a2}}\right)=0$
(b) $r_{b1}$ = 0.0424 m, $r_{b2}$ = 0.0300 m
$V_b=k\left(\frac{q_1}{r_{b1}}+\frac{q_2}{r_{b2}}\right)$
$V_b=(8.99 \times 10^9 \ N.m^2/C^2)\left(\frac{2.00 \times 10^{-6} \ C}{0.0424 \ m}+\frac{-2.00 \times 10^{-6} \ C}{0.0300 \ m}\right)$
$V_b=-1.75 \times 10^5$ V
(c) $W_{ab}=q_3(V_a-V_b)$
$W_{ab}=(-5.00 \times 10^{-6} \ C)[0-(-1.75 \times 10^5 \ V)]$
$W_{ab}=-0.875$ J
Since $V_b < V_a$ positive charge would be pulled by the existing charges from a to b, so they would do positive work on this charge. But they would repel a negative charge and hence do negative work on it, as we found in part (c).
Q#23.18
Two charges of equal magnitude Q are held a distance d apart. Consider only points on the line passing through both charges.
(a) If the two charges have the same sign, find the location of all points (if there are any) at which
(i) the potential (relative to infinity) is zero (is the electric field zero at these points?), and
(ii) the electric field is zero (is the potential zero at these points?).
(b) Repeat part (a) for two charges having opposite signs.
Answer:
The total potential is the scalar sum of the individual potentials, but the net electric field is the vector sum of the two fields.
The net potential can only be zero if one charge is positive and the other is negative, since it is a scalar. The electric field can only be zero if the two fields point in opposite directions.
(a) (i) Since both charges have the same sign, there are no points for which the potential is zero.
(ii) The two electric fields are in opposite directions only between the two charges, and midway between them the fields have equal magnitudes. So 0 E = midway between the charges, but V is never zero.
(b) (i) The two potentials have equal magnitude but opposite sign midway between the charges, so V = 0 midway between the charges, but E ≠ 0 there since the fields point in the same direction.
(ii) Between the two charges, the fields point in the same direction, so E cannot be zero there. In the other two regions, the field due to the nearer charge is always greater than the field due to the more distant charge, so they cannot cancel. Hence E is not zero anywhere.
It does not follow that the electric field is zero where the potential is zero, or that the potential is zero where the electric field is zero.
Q#23.19
Two point charges $q_1=2.40 \ nC$ and $q_1=-6.50 \ nC$ are 0.100 m apart. Point is midway between them; point B is 0.800 m from $q_1$ and 0.060 m from $q_2$ (Fig. E23.19).
Take the electric potential to be zero at infinity. Find
(a) the potential at point A
(b) the potential at point B
(c) the work done by the electric field on a charge of that travels from point B to point A.
Answer:
The locations of the changes and points A and B are sketched in Figure 23.19.
(a) $V_A=k\left(\frac{q_1}{r_{A1}}+\frac{q_2}{r_{A2}}\right)$
$V_b=(8.99 \times 10^9 \ N.m^2/C^2)\left(\frac{2.40 \times 10^{-9} \ C}{0.050 \ m}+\frac{-6.50 \times 10^{-9} \ C}{0.060 \ m}\right)$
$V_b=-737$ V
(b) $V_B=k\left(\frac{q_1}{r_{B1}}+\frac{q_2}{r_{B2}}\right)$
$V_b=(8.99 \times 10^9 \ N.m^2/C^2)\left(\frac{2.40 \times 10^{-9} \ C}{0.080 \ m}+\frac{-6.50 \times 10^{-9} \ C}{0.060 \ m}\right)$
(c) Use $W_{B→A}=q'(V_B-V_A)$ and the results of parts (a) and (b) to calculate W.
$W_{B→A}=(2.50 \times 10^{-9} \ C)(-704 \ V -(-737 \ V))$
$W_{B→A}=+8.2 \times 10^{-8}$ J
The electric force does positive work on the positive charge when it moves from higher potential (point B) to lower potential (point A).
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