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Chapter 7: Electric Potential

67 problems · 5 figures

Check Your Understanding 7.1 §7.1 source ↗

If Q has a mass of 4.00μg, what is the speed of Q at r2?

Answer

K= 1 2 m v 2 , v= 2 K m = 2 4.5 × 10 −7 J 4.00 × 10 −9 kg = 15m/s

Check Your Understanding 7.2 §7.1 source ↗

What is the potential energy of Q relative to the zero reference at infinity at r2 in the above example?

Answer

It has kinetic energy of 4.5×10−7J at point r2 and potential energy of 9.0×10−7J, which means that as Q approaches infinity, its kinetic energy totals three times the kinetic energy at r2, since all of the potential energy gets converted to kinetic.

Check Your Understanding 7.3 §7.1 source ↗

Is the electrical potential energy of two point charges positive or negative if the charges are of the same sign? Opposite signs? How does this relate to the work necessary to bring the charges into proximity from infinity?

Answer

positive, negative, and these quantities are the same as the work you would need to do to bring the charges in from infinity

Check Your Understanding 7.4 §7.2 source ↗

How much energy does a 1.5-V AAA battery have that can move 100 C?

Answer

ΔU= qΔV= (100C)(1.5V)= 150J

Check Your Understanding 7.5 §7.2 source ↗

How many electrons would go through a 24.0-W lamp each second from a 12-volt car battery?

Answer

–2.00 C, ne= 1.25×1019electrons

Check Your Understanding 7.6 §7.2 source ↗

How would this example change with a positron? A positron is identical to an electron except the charge is positive.

Answer

It would be going in the opposite direction, with no effect on the calculations as presented.

Check Your Understanding 7.7 §7.2 source ↗

From the examples, how does the energy of a lightning strike vary with the height of the clouds from the ground? Consider the cloud-ground system to be two parallel plates.

Answer

Given a fixed maximum electric field strength, the potential at which a strike occurs increases with increasing height above the ground. Hence, each electron will carry more energy. Determining if there is an effect on the total number of electrons lies in the future.

Check Your Understanding 7.8 §7.3 source ↗

What is the potential inside the metal sphere in Example 7.10?

Answer

V= kqr= (8.99×109N·m2/C2)(−3.00×10−9C5.00×10−3m)= −5390V; recall that the electric field inside a conductor is zero. Hence, any path from a point on the surface to any point in the interior will have an integrand of zero when calculating the change in potential, and thus the potential in the interior of the sphere is identical to that on the surface.

Check Your Understanding 7.9 §7.3 source ↗

What is the potential on the x-axis? The z-axis?

Answer

The x-axis the potential is zero, due to the equal and opposite charges the same distance from it. On the z-axis, we may superimpose the two potentials; we will find that for z>>d, again the potential goes to zero due to cancellation.

Check Your Understanding 7.10 §7.3 source ↗

What is the potential on the axis of a nonuniform ring of charge, where the charge density is λ(θ)=λcosθ ?

Answer

It will be zero, as at all points on the axis, there are equal and opposite charges equidistant from the point of interest. Note that this distribution will, in fact, have a dipole moment.

Check Your Understanding 7.11 §7.4 source ↗

Which coordinate system would you use to calculate the electric field of a dipole?

Answer

Any, but cylindrical is closest to the symmetry of a dipole.

Check Your Understanding 7.12 §7.5 source ↗

What are the equipotential surfaces for an infinite line charge?

Answer

infinite cylinders of constant radius, with the line charge as the axis

7-1 §7.1 source ↗

Would electric potential energy be meaningful if the electric field were not conservative?

Answer

No. We can only define potential energies for conservative fields.

7-3 §7.1 source ↗

Does the order in which we assemble a system of point charges affect the total work done?

Answer

No, though certain orderings may be simpler to compute.

7-5 §7.2 source ↗

What is the strength of the electric field in a region where the electric potential is constant?

Answer

The electric field strength is zero because electric potential differences are directly related to the field strength. If the potential difference is zero, then the field strength must also be zero.

7-7 §7.2 source ↗

Voltage is the common word for potential difference. Which term is more descriptive, voltage or potential difference?

Answer

Potential difference is more descriptive because it indicates that it is the difference between the electric potential of two points.

7-9 §7.2 source ↗

What is the relationship between voltage and energy? More precisely, what is the relationship between potential difference and electric potential energy?

Answer

They are very similar, but potential difference is a feature of the system; when a charge is introduced to the system, it will have a potential energy which may be calculated by multiplying the magnitude of the charge by the potential difference.

7-11 §7.2 source ↗

How are units of volts and electron-volts related? How do they differ?

Answer

An electron-volt is a volt multiplied by the charge of an electron. Volts measure potential difference, electron-volts are a unit of energy.

7-13 §7.3 source ↗

Compare the electric dipole moments of charges ±Q separated by a distance d and charges ±Q/2 separated by a distance d/2.

Answer

The second has 1/4 the dipole moment of the first.

7-15 §7.3 source ↗

In what region of space is the potential due to a uniformly charged sphere the same as that of a point charge? In what region does it differ from that of a point charge?

Answer

The region outside of the sphere will have a potential indistinguishable from a point charge; the interior of the sphere will have a different potential.

7-17 §7.4 source ↗

If the electric field is zero throughout a region, must the electric potential also be zero in that region?

Answer

No. It will be constant, but not necessarily zero.

7-19 §7.5 source ↗

If two points are at the same potential, are there any electric field lines connecting them?

Answer

no

7-21 §7.5 source ↗

Is the electric potential necessarily constant over the surface of a conductor?

Answer

No; it might not be at electrostatic equilibrium.

7-23 §7.5 source ↗

Can a positively charged conductor be at a negative potential? Explain.

Answer

Yes. It depends on where the zero reference for potential is. (Though this might be unusual.)

7-25 §7.6 source ↗

Why are the metal support rods for satellite network dishes generally grounded?

Answer

So that lightning striking them goes into the ground instead of the television equipment.

7-27 §7.6 source ↗

What are the similarities and differences between the processes in a photocopier and an electrostatic precipitator?

Answer

They both make use of static electricity to stick small particles to another surface. However, the precipitator has to charge a wide variety of particles, and is not designed to make sure they land in a particular place.

7-29 §7.1 source ↗

Consider a charge Q1(+5.0μC) fixed at a site with another charge Q2 (charge +3.0μC, mass 6.0μg) moving in the neighboring space. (a) Evaluate the potential energy of Q2 when it is 4.0 cm from Q1. (b) If Q2 starts from rest from a point 4.0 cm from Q1, what will be its speed when it is 8.0 cm from Q1? (Note: Q1 is held fixed in its place.)

Answer

a. U=3.4J;
b. 12mv2= kQ1Q2(1ri−1rf)→ v= 2.4x104m/s

7-31 §7.1 source ↗

To form a hydrogen atom, a proton is fixed at a point and an electron is brought from far away to a distance of 0.529×10−10m, the average distance between proton and electron in a hydrogen atom. How much work is done?

Answer

U= 4.36× 10 −18 J

7-33 §7.2 source ↗

Find the ratio of speeds of an electron and a negative hydrogen ion (one having an extra electron) accelerated through the same voltage, assuming non-relativistic final speeds. Take the mass of the hydrogen ion to be 1.67×10−27kg.

Answer

1 2 m e v e 2 = q V , 1 2 m H v H 2 = q V , so that m e v e 2 m H v H 2 = 1 or v e v H = 42.8

7-35 §7.2 source ↗

Show that units of V/m and N/C for electric field strength are indeed equivalent.

Answer

1V= 1J/C;1J= 1N·m→ 1V/m= 1N/C

7-37 §7.2 source ↗

The electric field strength between two parallel conducting plates separated by 4.00 cm is 7.50×104V/m. (a) What is the potential difference between the plates? (b) The plate with the lowest potential is taken to be zero volts. What is the potential 1.00 cm from that plate and 3.00 cm from the other?

Answer

a. VAB=3.00kV; b. VAB=750V

7-39 §7.2 source ↗

Two parallel conducting plates are separated by 10.0 cm, and one of them is taken to be at zero volts. (a) What is the electric field strength between them, if the potential 8.00 cm from the zero volt plate (and 2.00 cm from the other) is 450 V? (b) What is the voltage between the plates?

Answer

a. VAB= Ed→ E= 5.63kV/m;
b. VAB=563V

7-41 §7.2 source ↗

An electron is to be accelerated in a uniform electric field having a strength of 2.00×106V/m. (a) What energy in keV is given to the electron if it is accelerated through 0.400 m? (b) Over what distance would it have to be accelerated to increase its energy by 50.0 GeV?

Answer

a. ΔK=qΔVandVAB=Ed,so thatΔK=800keV;
b. d=25.0km

7-43 §7.2 source ↗

The electric field in a region is pointed away from the z-axis and the magnitude depends upon the distance s from the axis. The magnitude of the electric field is given as E=αs where α is a constant. Find the potential difference between points P1andP2, explicitly stating the path over which you conduct the integration for the line integral.

The figure shows two points P subscript 1 and P subscript 2 at distances a and b from the origin and having an angle phi between them.

Answer

One possibility is to stay at constant radius and go along the arc from P1 to P2, which will have zero potential due to the path being perpendicular to the electric field. Then integrate from a to b: Vab=αln(ab) The potential at point P1 is higher than at P2 since the field points radially outward, and this is consistent with a positive potential difference going from distance a to a larger distance b.

7-45 §7.3 source ↗

A 0.500-cm-diameter plastic sphere, used in a static electricity demonstration, has a uniformly distributed 40.0-pC charge on its surface. What is the potential near its surface?

Answer

V = 144 V

7-47 §7.3 source ↗

If the potential due to a point charge is 5.00×102V at a distance of 15.0 m, what are the sign and magnitude of the charge?

Answer

V= kQr→ Q= 8.33×10−7C;
The charge is positive because the potential is positive.

7-49 §7.3 source ↗

A research Van de Graaff generator has a 2.00-m-diameter metal sphere with a charge of 5.00 mC on it. Assume the potential energy is zero at a reference point infinitely far away from the Van de Graaff. (a) What is the potential near its surface? (b) At what distance from its center is the potential 1.00 MV? (c) An oxygen atom with three missing electrons is released near the Van de Graaff generator. What is its kinetic energy in MeV when the atom is at the distance found in part b?

Answer

a. V=45.0MV;
b. V= kQr→ r= 45.0m;
c. ΔU=132MeV

7-51 §7.3 source ↗

(a) What is the potential between two points situated 10 cm and 20 cm from a 3.0-μC point charge? (b) To what location should the point at 20 cm be moved to increase this potential difference by a factor of two?

Answer

V=kQ/r; a. Relative to origin, find the potential at each point and then calculate the difference.
ΔV=135×103V;
b. To double the potential difference, move the point from 20 cm to infinity; the potential at 20 cm is halfway between zero and that at 10 cm.

7-53 §7.3 source ↗

Two charges –2.0µCand+2.0µC are separated by 4.0 cm on the z-axis symmetrically about origin, with the positive one uppermost. Two space points of interest P1andP2 are located 3.0 cm and 30 cm from origin at an angle 30° with respect to the z-axis. Evaluate electric potentials at P1andP2 in two ways: (a) Using the exact formula for point charges, and (b) using the approximate dipole potential formula.

Answer

a. VP1= 7.4×105V
and VP2= 6.9×103V;
b. VP1= 6.9×105V and VP2= 6.9×103V

7-55 §7.4 source ↗

Throughout a region, equipotential surfaces are given by z=constant. The surfaces are equally spaced with V=100V for z= 0.00m,V= 200V for z= 0.50m,V= 300V for z=1.00m. What is the electric field in this region?

Answer

The problem is describing a uniform field, so E=200V/m in the –z-direction.

7-57 §7.4 source ↗

Calculate the electric field of an infinite line charge, throughout space.

Answer

Apply E→=−∇→V with ∇→= r^∂∂r+ φ^1r∂∂φ+ z^∂∂z to the potential calculated earlier,
V= −2kλlns:E→= 2kλ1rr^ as expected.

7-59 §7.5 source ↗

A very large sheet of insulating material has had an excess of electrons placed on it to a surface charge density of –3.00nC/m2. (a) As the distance from the sheet increases, does the potential increase or decrease? Can you explain why without any calculations? Does the location of your reference point matter? (b) What is the shape of the equipotential surfaces? (c) What is the spacing between surfaces that differ by 1.00 V?

Answer

a. increases; the constant (negative) electric field has this effect, the reference point only matters for magnitude; b. they are planes parallel to the sheet; c. 0.006 m/V

7-61 §7.5 source ↗

Two large charged plates of charge density ±30μC/m2 face each other at a separation of 5.0 mm. (a) Find the electric potential everywhere. (b) An electron is released from rest at the negative plate; with what speed will it strike the positive plate?

Answer

a. from the previous chapter, the electric field has magnitude σε0 in the region between the plates and zero outside; defining the negatively charged plate to be at the origin and zero potential, with the positively charged plate located at +5mm in the z-direction, V=1.7×104V so the potential is 0 for z< 0, 1.7×104V(z5mm) for 0≤z≤5mm,1.7×104V for z>5mm;
b. qV= 12mv2→ v= 7.7×107m/s

7-63 §7.5 source ↗

Two parallel plates 10 cm on a side are given equal and opposite charges of magnitude 5.0×10−9C. The plates are 1.5 mm apart. What is the potential difference between the plates?

Answer

V = 85 V

7-65 §7.5 source ↗

Concentric conducting spherical shells carry charges Q and –Q, respectively. The inner shell has negligible thickness. What is the potential difference between the shells?

The figure shows two concentric spheres. The inner sphere has radius a and charge Q. The outer sphere is a shell with inner radius b and outer radius c and charge -Q.

Answer

In the region a≤r≤b, E→= kQr2r^, and E is zero elsewhere; hence, the potential difference is V=kQ(1a−1b).

7-67 §7.5 source ↗

A solid cylindrical conductor of radius a is surrounded by a concentric cylindrical shell of inner radius b. The solid cylinder and the shell carry charges Q and –Q, respectively. Assuming that the length L of both conductors is much greater than a or b, what is the potential difference between the two conductors?

Answer

From previous results VP− VR= −2kλlnsPsR., note that b is a very convenient location to define the zero level of potential: ΔV= −2kQLlnab.

7-69 §7.6 source ↗

(a) What is the direction and magnitude of an electric field that supports the weight of a free electron near the surface of Earth? (b) Discuss what the small value for this field implies regarding the relative strength of the gravitational and electrostatic forces.

Answer

a. F= 5.58×10−11N/C;

The electric field is towards the surface of Earth. b. The coulomb force is much stronger than gravity.

7-71 §7.6 source ↗

In a Geiger counter, a thin metallic wire at the center of a metallic tube is kept at a high voltage with respect to the metal tube. Ionizing radiation entering the tube knocks electrons off gas molecules or sides of the tube that then accelerate towards the center wire, knocking off even more electrons. This process eventually leads to an avalanche that is detectable as a current. A particular Geiger counter has a tube of radius R and the inner wire of radius a is at a potential of V0 volts with respect to the outer metal tube. Consider a point P at a distance s from the center wire and far away from the ends. (a) Find a formula for the electric field at a point P inside using the infinite wire approximation. (b) Find a formula for the electric potential at a point P inside. (c) Use V0= 900V, a= 3.00mm,R= 2.00cm, and find the value of the electric field at a point 1.00 cm from the center.

The figure shows the schematic of a Geiger counter.

Answer

We know from the Gauss’s law chapter that the electric field for an infinite line charge is E→P=2kλ1ss^, and from earlier in this chapter that the potential of a wire-cylinder system of this sort is VP= −2kλlnsPR by integration. We are not given λ, but we are given a fixed V0; thus, we know that V0=−2kλlnaR and hence λ= − V02kln(aR). We may substitute this back in to find a. E→P= − V0ln(aR)1ss^; b. VP= V0ln(sPR)ln(aR); c. 4.74×104N/C

7-73 §7.6 source ↗

To form a helium atom, an alpha particle that contains two protons and two neutrons is fixed at one location, and two electrons are brought in from far away, one at a time. The first electron is placed at 0.600×10−10m from the alpha particle and held there while the second electron is brought to 0.600×10−10m from the alpha particle on the other side from the first electron. See the final configuration below. (a) How much work is done in each step? (b) What is the electrostatic energy of the alpha particle and two electrons in the final configuration?

The figure shows alpha particle with electrons on left and right sides at distance 0.6 times 10 super script -10 meters.

Answer

a. U1=7.68×10−18JU2=5.76×10−18J;
b. U1+ U2= −1.34×10−17J

7-75 §7.6 source ↗

The probability of fusion occurring is greatly enhanced when appropriate nuclei are brought close together, but mutual Coulomb repulsion must be overcome. This can be done using the kinetic energy of high-temperature gas ions or by accelerating the nuclei toward one another. (a) Calculate the potential energy of two singly charged nuclei separated by 1.00×10−12m. (b) At what temperature will atoms of a gas have an average kinetic energy equal to this needed electrical potential energy?

Answer

a. U= 2.30×10−16J;
b. K−= 32kT→ T= 1.11×107K

7-77 §7.6 source ↗

An electron enters a region between two large parallel plates made of aluminum separated by a distance of 2.0 cm and kept at a potential difference of 200 V. The electron enters through a small hole in the negative plate and moves toward the positive plate. At the time the electron is near the negative plate, its speed is 4.0×105m/s. Assume the electric field between the plates to be uniform, and find the speed of electron at (a) 0.10 cm, (b) 0.50 cm, (c) 1.0 cm, and (d) 1.5 cm from the negative plate, and (e) immediately before it hits the positive plate.

The figure shows two charged parallel plates – one positive and one negative and an electron entering between the plates. The distance between the plates is 2cm and the potential difference is 200V.

Answer

a. 1.9×106m/s; b. 4.2×106m/s; c. 5.9×106m/s; d. 7.3×106m/s; e. 8.4×106m/s

7-79 §7.6 source ↗

(a) Will the electric field strength between two parallel conducting plates exceed the breakdown strength of dry air, which is 3.00×106V/m, if the plates are separated by 2.00 mm and a potential difference of 5.0×103V is applied? (b) How close together can the plates be with this applied voltage?

Answer

a. E=2.5×106V/m<3×106V/mNo, the field strength is smaller than the breakdown strength for air.;
b. d=1.7mm

7-81 §7.6 source ↗

A double charged ion is accelerated to an energy of 32.0 keV by the electric field between two parallel conducting plates separated by 2.00 cm. What is the electric field strength between the plates?

Answer

K f = q V AB = q E d → E = 8.00 × 10 5 V/m

7-83 §7.6 source ↗

A lightning bolt strikes a tree, moving 20.0 C of charge through a potential difference of 1.00×102MV. (a) What energy was dissipated? (b) What mass of water could be raised from 15°C to the boiling point and then boiled by this energy? (c) Discuss the damage that could be caused to the tree by the expansion of the boiling steam.

Answer

a. Energy= 2.00×109J;
b. Q=m(cΔT+L∇)m=766kg;
c. The expansion of the steam upon boiling can literally blow the tree apart.

7-85 §7.6 source ↗

(a) A sphere has a surface uniformly charged with 1.00 C. At what distance from its center is the potential 5.00 MV? (b) What does your answer imply about the practical aspect of isolating such a large charge?

Answer

a. V= kQr→ r= 1.80km; b. A 1-C charge is a very large amount of charge; a sphere of 1.80 km is impractical.

7-87 §7.6 source ↗

In one of the classic nuclear physics experiments at the beginning of the twentieth century, an alpha particle was accelerated toward a gold nucleus, and its path was substantially deflected by the Coulomb interaction. If the energy of the doubly charged alpha nucleus was 5.00 MeV, how close to the gold nucleus (79 protons) could it come before being deflected?

Answer

The alpha particle approaches the gold nucleus until its original energy is converted to potential energy. 5.00MeV= 8.00×10−13J, so
E0=qkQr→r=4.54×10−14m
(Size of gold nucleus is about 7×10−15m).

7-89 §7.AP source ↗

A battery-operated car uses a 12.0-V system. Find the charge the batteries must be able to move in order to accelerate the 750 kg car from rest to 25.0 m/s, make it climb a 2.00×102-m high hill, and finally cause it to travel at a constant 25.0 m/s while climbing with 5.00×102-N force for an hour.

Answer

E tot = 4.67 × 10 7 J E tot = q V → q = E tot V = 3.89 × 10 6 C

7-91 §7.AP source ↗

A uniformly charged half-ring of radius 10 cm is placed on a nonconducting table. It is found that 3.0 cm above the center of the half-ring the potential is –3.0 V with respect to zero potential at infinity. How much charge is in the half-ring?

Answer

V P = k q tot z 2 + R 2 → q tot = −3.5× 10 −11 C

7-93 §7.AP source ↗

A CD disk of radius (R=3.0cm) is sprayed with a charged paint so that the charge varies continually with radial distance r from the center in the following manner: σ= − (6.0C/m)r/R.

Find the potential at a point 4 cm above the center.

Answer

V P = −2.2 GV

7-95 §7.AP source ↗

A large metal plate is charged uniformly to a density of σ= 2.0×10−9C/m2. How far apart are the equipotential surfaces that represent a potential difference of 25 V?

Answer

Recall from the previous chapter that the electric field EP=σ02ε0 is uniform throughout space, and that for uniform fields we have E=−ΔVΔz for the relation. Thus, we get σ2ε0= ΔVΔz→ Δz= 0.22m for the distance between 25-V equipotentials.

7-97 §7.AP source ↗

(a) Find x>>L limit of the potential of a finite uniformly charged rod and show that it coincides with that of a point charge formula. (b) Why would you expect this result?

Answer

a. Take the result from Example 7.13, divide both the numerator and the denominator by x, take the limit of that, and then apply a Taylor expansion to the resulting log to get: VP≈kλLx; b. which is the result we expect, because at great distances, this should look like a point charge of q=λL
Example 7.13 Potential of a Line of Charge in the book ↗

7-99 §7.AP source ↗

Two parallel conducting plates, each of cross-sectional area 400cm2, are 2.0 cm apart and uncharged. If 1.0×1012 electrons are transferred from one plate to the other, (a) what is the potential difference between the plates? (b) What is the potential difference between the positive plate and a point 1.25 cm from it that is between the plates?

Answer

a. V=9.0×103V; b. −9.0×103V(1.25cm2.0cm)= −5.7×103V

7-101 §7.AP source ↗

Earth has a net charge that produces an electric field of approximately 150 N/C downward at its surface. (a) What is the magnitude and sign of the excess charge, noting the electric field of a conducting sphere is equivalent to a point charge at its center? (b) What acceleration will the field produce on a free electron near Earth’s surface? (c) What mass object with a single extra electron will have its weight supported by this field?

Answer

a. E= KQr2→ Q= −6.76×105C;
b. F=ma=qE→a=qEm=2.63×1013m/s2(upwards);
c. F= −mg= qE→ m= −qEg= 2.45×10−18kg

7-103 §7.AP source ↗

What can you say about two charges q1andq2, if the electric field one-fourth of the way from q1toq2 is zero?

Answer

If the electric field is zero ¼ from the way of q1andq2, then we know from
E= kQr2that|E1|= |E2|→ Kq1x2= Kq2(3x)2so thatq2q1= (3x)2x2= 9; the charge q2 is 9 times larger than q1.

7-105 §7.AP source ↗

An electron has an initial velocity of 5.00×106m/s in a uniform 2.00×105-N/C electric field. The field accelerates the electron in the direction opposite to its initial velocity. (a) What is the direction of the electric field? (b) How far does the electron travel before coming to rest? (c) How long does it take the electron to come to rest? (d) What is the electron’s velocity when it returns to its starting point?

Answer

a. The field is in the direction of the electron’s initial velocity.
b. v2=v02+2ax→x=−v022a(v=0).Also,F=ma=qE→a=qEm,x=3.56×10−4m;
c. v2=v0+at→t=−v0mqE(v=0),∴t=1.42×10−10s;
d. v= −(2qExm)1/2−5.00×106m/s(opposite its initial velocity)

7-107 §7.CP source ↗

Look up (presumably online, or by dismantling an old device and making measurements) the magnitude of the potential deflection plates (and the space between them) in an ink jet printer. Then look up the speed with which the ink comes out the nozzle. Can you calculate the typical mass of an ink drop?

Answer

Answers will vary. This appears to be proprietary information, and ridiculously difficult to find. Speeds will be 20 m/s or less, and there are claims of ~10−7 grams for the mass of a drop.

7-109 §7.CP source ↗

Calculate the electric field of a dipole throughout space from the potential.

Answer

Apply E→=−∇→V with ∇→= r^∂∂r+ θ^1r∂∂θ+ φ^1rsinθ∂∂φ to the potential calculated earlier, VP=kp→·r^r2 with p→=qd→, and assume that the axis of the dipole is aligned with the z-axis of the coordinate system. Thus, the potential is VP= kqd→·r^r2= kqdcosθr2.
E→= 2kqd(cosθr3)r^+ kqd(sinθr3)θ^

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Problems and answers are adapted from University Physics Volume 2 by Samuel J. Ling, Jeff Sanny and William Moebs, OpenStax, and are used under the CC BY-NC-SA 4.0 licence. Access for free at https://openstax.org/books/university-physics-volume-2/pages/1-introduction. Illustrations: Copyright Rice University, OpenStax, under CC BY-NC-SA 4.0 license. This adaptation is released under the same licence, is non-commercial, and is neither affiliated with nor endorsed by OpenStax or Rice University.