19.4 Exercises 653 (13) Silver has 47 electrons per atom and a molar mass of 107.87 kg/kmol An electrically neutral pin of silver has a mass of 10 g (a) Calculate the number of electrons in the silver pin (Avogadro’s number is 6.022 × 1026 atoms/kmol) (b) Electrons are added to the pin until the net charge is −1 mC How many electrons are added for every billion (109 ) electrons in the neutral atoms? (14) Two protons in an atomic nucleus are separated by × 10−15 m (a typical internuclear dimension) (a) Find the magnitude of the electrostatic repulsive force between the protons (b) How does the magnitude of the electrostatic force compare to the magnitude of the gravitational force between the two protons? (15) Two particles have an identical charge q and an identical mass m What must the charge-mass ratio, q/m, of the two particles be if the magnitude of their electrostatic force equals the magnitude of the gravitational force (16) Two equally charged pith balls are at a distance r = cm apart, as shown in Fig 19.17 Find the magnitude of the charge on each ball if they repel each other with a force of magnitude × 10−5 N Does the answer give you any hint about the exact sign of each charge? Explain Fig 19.17 See Exercise (16) q q r (17) Two point charges q1 and q2 are m apart, and their combined charge is 40 μC (a) If one repels the other with a force of 0.175 N, what are the two charges? (b) If one attracts the other with a force of 0.225 N, what are the two charges? (18) Two point charges q1 = +4 μC and q2 = +6 μC are 10 cm apart A point charge q3 = +2 μC is placed midway between q1 and q2 Find the magnitude and direction of the resultant force on q3 (19) Three μC point charges are placed along a straight line as shown in Fig 19.18 Calculate the net force on each charge (20) Three point charges q1 = q2 = q3 = −4 μC are located at the corners of an equilateral triangle as shown in Fig 19.19 (a) Calculate the magnitude of 654 19 Electric Force the net force on any one of the three charges (b) If the charges are positive, i.e q1 = q2 = q3 = +4 μC, would this change the magnitude calculated in part a? Fig 19.18 See Exercise (19) q =+ C q =+ C q =+ C + + + 3m 3m Fig 19.19 See Exercise (20) q1 – 1m 1m – q – 1m q (21) Three point charges q1 = +2 μC, q2 = −3 μC, and q3 = + μC are located at the corners of a right angle triangle as shown in Fig 19.20 Find the magnitude and direction of the resultant force on q3 Fig 19.20 See Exercise (21) q1 + y x 5cm q2 – 30° + q3 (22) Three equal point charges of magnitude q lie on a semicircle of radius R as √ shown in Fig 19.21 Show that the net force on q2 has a magnitude kq2 / 2R2 and points downward away from the center C of the semicircle (23) Four equal point charges, q1 = q2 = q3 = q4 = +3 μC, are placed at the four corners of a square that has a side a = 0.4 m, see Fig 19.22 (a) Find the force 19.4 Exercises 655 on q1 (b) Find the force exerted on a test charge of C placed at the center P of the square Fig 19.21 See Exercise (22) q1 q3 R C R q2 Fig 19.22 See Exercise (23) q1 + + q4 y x P q2 + a = 0.4m + q (24) Four equal point charges q1 = q2 = q3 = q4 = −1 μC, are located as shown in Fig 19.23 (a) Calculate the net force exerted on the charge q4 , which is located midway between q1 and q3 (b) Calculate the magnitude and direction of the net force on the charge q2 Fig 19.23 See Exercise (24) q1 – y x – 1m q2 – 1m q4 – q3 (25) A negative point charge of magnitude q is located on the x-axis at point x = −a, and a positive point charge of the same magnitude is located at x = +a, see Fig 19.24 A third positive point charge q◦ is located on the y-axis with a coordinate 656 19 Electric Force (0, y) (a) What is the magnitude and direction of the force exerted on q◦ when it is at the origin (0,0)? (b) What is the force on q◦ when its coordinate is (0, y)? (c) Sketch a graph of the force on q◦ as a function of y, for values of y between −4a and +4a Fig 19.24 See Exercise (25) y + + qo y r− -q – θ r+ θ +q + (-a,0 ) x (a,0 ) (26) In the Bohr model of the hydrogen atom, an electron of mass m = 9.11 × 10−31 kg revolves about a stationary proton in a circular orbit of radius r = 5.29 × 10−11 m, see Fig 19.25 (a) What is the magnitude of the electrical force on the electron? (b) What is the magnitude of the centripetal acceleration of the electron? (c) What is the orbital speed of the electron? Fig 19.25 See Exercise (26) Electron r Fe Fe Proton (27) In the cesium chloride crystal (CsCl), eight Cs+ ions are located at the corners of a cube of side a = 0.4 nm and a Cl− ion is at the center, see Fig 19.26 What is the magnitude of the electrostatic force exerted on the Cl− ion by: (a) the eight Cs+ ions?, (b) only seven Cs+ ions? (28) Two positive point charges q1 and q2 are set apart by a fixed distance d and have a sum Q = q1 + q2 For what values of the two charges is the Coulomb force maximum between them? 19.4 Exercises 657 Fig 19.26 See Exercise (27) a Cs Cl (29) Two equal positive charges q are held stationary on the x-axis, one at x = − a and the other at x = +a A third charge +q of mass m is in equilibrium at x = and constrained to move only along the x-axis The charge +q is then displaced from the origin to a small distance x a and released, see Fig 19.27 (a) Show that +q will execute a simple harmonic motion and find an expression for its period T (b) If all three charges are singly-ionized atoms (q = q = +e) each of mass m = 3.8 × 10−25 kg and a = × 10−10 m, find the oscillation period T Fig 19.27 See Exercise (29) + q′ +q +q x a a (30) Two identical small spheres of mass m and charge q hang from non-conducting strings, each of length L At equilibrium, each string makes an angle θ with the vertical, see Fig 19.28 (a) When θ is so small that tan θ sin θ , show that the separation distance r between the spheres is r = (Lq2 /2π ◦ mg)1/3 (b) If L = 10 cm, m = g, and r = 1.7 cm, what is the value of q? Fig 19.28 See Exercise (30) θ L q L θ x x r q 658 19 Electric Force (31) For the charge distribution shown in Fig 19.29, the long non-conducting massless rod of length L (which is pivoted at its center) is balanced horizontally when a weight W is placed at a distance x from the center (a) Find the distance x and the force exerted by the rod on the pivot (b) What is the value of h when the rod exerts no force on the pivot? Fig 19.29 See Exercise (31) L x +q +2q h W Pivot h +Q +Q (32) Two small charged spheres hang from threads of equal length L The first sphere has a positive charge q, mass m, and makes a small angle θ1 with the vertical, while the second sphere has a positive charge q, mass 3m, and makes a smaller angle θ2 with the vertical, see Fig 19.30 For small angles, take tan θ θ and assume that the spheres only have horizontal displacements and hence the electric force of repulsion is always horizontal (a) Find the ratio θ1 /θ2 (b) Find the distance r between the spheres Fig 19.30 See Exercise (32) L T1 L θ1 θ2 m T2 3m q 2q r 20 Electric Fields In this chapter, we introduce the concept of an electric field associated with a variety of charge distributions We follow that by introducing the concept of an electric field in terms of Faraday’s electric field lines In addition, we study the motion of a charged particle in a uniform electric field 20.1 The Electric Field Based on the electric force between charged objects, the concept of an electric field was developed by Michael Faraday in the 19th century, and has proven to have valuable uses as we shall see In this approach, an electric field is said to exist in the region of space around → any charged object To visualize this assume an electrical force of repulsion F between two positive charges q (called source charge) and q◦ (called test charge), see Fig 20.1a Now, let the charge q◦ be removed from point P where it was formally located as → shown in Fig 20.1b The charge q is said to set up an electric field E at P, and if q◦ → is now placed at P, then a force F is exerted on q◦ by the field rather than by q, see Fig 20.1c Since force is a vector quantity, the electric field is a vector whose properties are determined from both the magnitude and the direction of an electric force We define → the electric field vector E as follows: H A Radi and J O Rasmussen, Principles of Physics, Undergraduate Lecture Notes in Physics, DOI: 10.1007/978-3-642-23026-4_20, © Springer-Verlag Berlin Heidelberg 2013 659 660 20 Electric Fields Spotlight → The electric field vector E at a point in space is defined as the electric → force F acting on a positive test charge q◦ located at that point divided by the magnitude of the test charge: → → E = F q◦ Fig 20.1 (a) A charge q (20.1) q → + exerts a force F on a test +++ + (a) charge q◦ at point P (b) The → electric field E established at qo + F P P due to the presence of q (c) → → The force F = q◦ E exerted q → by E on the test charge q◦ + +++ + (b) q + +++ + (c) P E qo + P F = qo E This equation can be rearranged as follows (see Fig 20.1c): → → F = q◦ E (20.2) → The SI unit of the electric field E is newton per coulomb (N/C) → The direction of E is the direction of the force on a positive test charge placed in the field, see Fig 20.2 20.2 The Electric Field of a Point Charge To find the magnitude and direction of an electric field, we consider a positive point charge q as a source charge A positive test charge q◦ is then placed at point P, 20.2 The Electric Field of a Point Charge 661 a distance r away from q, see Fig 20.3 From Coulomb’s law, the force exerted on q◦ is: → F =k q q◦ → rˆ r2 (20.3) q (a) + +++ + E P - - q (b) E P → Fig 20.2 (a) If the charge q is positive, then the force F on the test charge q◦ (not shown in the figure) → at point P is directed away from q Therefore, the electric field E at P is directed away from q (b) If the → charge q is negative, then the force F on q◦ at point P is directed toward q Therefore, the electric field → E at P is directed toward q where → rˆ is a unit vector directed from the source charge q to the test charge q◦ This force has the same direction as the unit vector → rˆ → → Since the electric field at point P is defined from Eq 20.1 as E = F /q◦ , then according to Fig 20.3, the electric field created at P by q is an outward vector given by: → E =k qo r + (20.4) E F r P r q q→ rˆ r2 + P r q + → Fig 20.3 If the point charge q is positive, then both the force F on the positive test charge q◦ and the → electric field E at point P are directed away from q 662 20 Electric Fields → → When the source charge q is negative, the force F on q◦ and the electric field E at point P will be toward q, see Fig 20.4 Note that for both positive and negative charges, → rˆ is a unit vector that is always directed from the source charge q to the point P, see Figs 20.3 and 20.4 In all previous and coming discussions, the positive test charge q◦ must be very small, so that it does not disturb the charge distribution of the source charge q → Mathematically, this can be done by taking the limit of the ratio F /q◦ when q◦ approaches zero Thus: → → F q◦ →0 q◦ E = lim (20.5) qo F + E r P r r q P r q - - → Fig 20.4 If the point charge q is negative, then both the force F on the test positive charge q◦ and the → electric field E at point P are directed toward q, but the unit vector → rˆ remains pointed toward P The electric field due to a group of point charges q1 , q2 , q3 at point P can be obtained by first using Eq 20.4 to calculate the electric field of each individual charge, such that: → En = k qn → rˆn rn2 (n = 1, 2, 3, ) (20.6) → Then we calculate the vector sum E of the electric fields of all the charges This sum is expressed as follows: → → → → E = E1 + E2 + E3 + = k n qn → rˆn rn2 (n = 1, 2, 3, ) (20.7) 20.2 The Electric Field of a Point Charge 663 where rn is the distance from the nth source charge qn to the point P and → rˆn is a unit vector directed away from qn to P It is clear that Eq 20.7 exhibits the application of the superposition principle to electric fields Example 20.1 √ Four point charges q1 = q2 = Q and q3 = q4 = −Q, where Q = μC, are placed at the four corners of a square of side a = 0.4 m, see Fig 20.5a Find the electric field at the center P of the square a + q1 - q4 a + q1 - q4 E2 a q2 45 ° a P - + q3 q2 P + E 45° E13 - x q3 (b) (a) Fig 20.5 Solution: The distance between each charge and the center P of the square is √ a/ At point P, the point charges q1 and q3 produce two diagonal electric field → → vectors E and E , both directed toward q3 , see Fig 20.5b Hence, their vector → → → sum E 13 = E + E points toward q3 and has the magnitude: E13 = E1 + E3 = k Q Q Q +k = 4k √ √ a (a/ 2)2 (a/ 2)2 → → At point P, the charges q2 and q4 produce two diagonal electric fields E and E , → → → both directed toward q4 , see Fig 20.5b Hence, their vector sum E 24 = E + E points toward q4 and has the magnitude: E24 = E2 + E4 = k Q Q Q +k = 4k √ √ 2 a (a/ 2) (a/ 2) → → We now must combine the two electric field vectors E 13 and E 24 to form → → → the resultant electric field vector E = E 13 + E 24 which is along the positive x-direction and has the magnitude: 664 20 Electric Fields Q 8Q E = E13 cos 45◦ + E24 cos 45◦ = × 4k × √ = k√ a 2a2 √ −6 8( × 10 m) = (9 × 109 N.m2 /C2 ) √ = 4.5 × 105 N/C (0.4 m)2 Example 20.2 (Electric Dipole) Consider two point charges q1 = −24 nC and q2 = +24 nC that are 10 cm apart, forming an electric dipole, see Fig 20.6 Calculate the electric field due to the two charges at points a, b, and c Fig 20.6 E 2c Ec c E 1c 10 cm q1 10 cm 60° Ea a cm q2 cm b Eb cm Solution: At point a, the electric field vector due to the negative charge q1 , is directed toward the left, and its magnitude is: E1a = k −9 C) |q1 | 2 (24 × 10 = (9 × 10 N.m /C ) = 135 × 103 N/C (0.04 m)2 r1a The electric field vector due to the positive charge q2 is also directed toward the left, and its magnitude is: E2a = k −9 C) |q2 | 2 (24 × 10 = (9 × 10 N.m /C ) = 60 × 103 N/C (0.06 m)2 r2a Then, the resultant electric field at point a is toward the left and its magnitude is: Ea = E1a + E2a = 135 × 103 N/C + 60 × 103 N/C = 195 × 103 N/C (Toward the left) 20.2 The Electric Field of a Point Charge 665 At point b, the electric field vector due to the negative charge q1 , is directed toward the left, and its magnitude is: E1b = k |q1 | (24 × 10−9 C) = (9 × 109 N.m2 /C2 ) = 15 × 103 N/C (0.12 m)2 r1b In addition, the electric field vector due to the positive charge q2 is directed toward the right, and its magnitude is: E2b = k −9 C) |q2 | 2 (24 × 10 = (9 × 10 N.m /C ) = 540 × 103 N/C (0.02 m)2 r2b Since E2b > E1b , the resultant electric field at point b is toward the right and its magnitude is: Eb = E2b − E1b = 540 × 103 N/C − 15 × 103 N/C = 525 × 103 N/C (Toward the right) → → At point c, the magnitudes of the electric field vectors E 1c and E 2c established by q1 and q2 are the same because |q1 | = |q2 | = 24 nC and r1c = r2c = 10 cm Thus: E2c = E1c = k −9 C) |q1 | 2 (24 × 10 = (9 × 10 N.m /C ) = 21.6 × 103 N/C (0.1 m)2 r1c The triangle formed from q1 , q2 , and point c in Fig 20.6 is an equilateral triangle of angle 60◦ Hence, from geometry, the vertical components of the → → two vectors E 1c and E 2c cancel each other The horizontal components are both directed toward the left and add up to give the resultant electric field Ec at point c, see the figure below E 2c Ec E 1c cos60° E 2c cos60° 60° c 60° E 1c Thus: Ec = E1c cos 60◦ + E2c cos 60◦ = 2E1c cos 60◦ = 2(21.6 × 103 N/C)(0.5) = 21.6 × 103 N/C (Toward the left) 666 20.3 20 Electric Fields The Electric Field of an Electric Dipole Generally, the electric dipole introduced in Example 20.3 consists of a positive charge q+ = +q and a negative charge q− = −q separated by a distance 2a, see Fig 20.7 In this figure, the dipole axis is taken to be along the x-axis and the origin of the xy plane is taken to be at the center of the dipole Therefore, the coordinates of q+ and q− are (+a, 0) and (−a, 0), respectively y E E+ P (x,y) E− r− -q r+ +q r− - (-a,0) (a,0) y r+ + x x-a x+a → → → Fig 20.7 The electric field E = E + + E − at point P(x, y) due to an electric dipole located along the x-axis The dipole has a length 2a Let us assume that a point P(x, y) exists in the xy-plane as shown in Fig 20.7 We → will call the electric field produced by the positive charge E+ and the electric field → produced by the negative charge E− Using the superposition principle, the total electric field at P is: → → → E = E+ + E− = k q+ → q− ˆ rˆ + k → r− + r+ r− (20.8) = (x − a)2 + y2 and r = (x + a)2 + y2 From the geometry of Fig 20.7, we have r+ − In addition, → rˆ+ is a unit vector directed outwards and away from the positive charge q at (+a, 0) On the other hand, → rˆ is a unit vector directed outwards and away + − from the negative charge q− at (−a, 0) Accordingly, Eq 20.8 becomes: → E =k −q q → → rˆ+ + rˆ− (x − a)2 + y2 (x + a)2 + y2 (20.9) 20.3 The Electric Field of an Electric Dipole 667 Therefore, the general electric field will take the following form: → E = kq (x → rˆ+ − a)2 + y2 − (x → rˆ− + a)2 (20.10) + y2 The Electric Field Along the Dipole Axis Let us first assume a point P exists on the dipole axis, i.e y = 0, and satisfies the → → condition x < −a, as shown in Fig 20.8a In this case, → rˆ+ = → rˆ− = − i , where i is a unit vector along the x-axis E+ P E− -q (a) x < −a +q -a -q (b) E+ E− P - − a< x< + a a (c) -q → + x +a +q -a → 0 x +a +q - + + +a E− P E+ x x > +a → Fig 20.8 The electric field E = E+ + E− at different points along the axis of a dipole that has a length 2a When P has an x-coordinate that satisfies −a < x < + a as in Fig 20.8b, then → → → rˆ+ = − i and → rˆ− = + i When P satisfies x > + a as in Fig 20.8c, then → rˆ+ = → → rˆ = + i Substituting in Eq 20.10, we get: − ⎧ 1 ⎪ ⎪ −kq − ⎪ ⎪ (x − a) (x + a)2 ⎪ ⎪ ⎨ → 1 + E = −kq ⎪ (x − a) (x + a)2 ⎪ ⎪ ⎪ ⎪ 1 ⎪ ⎩ kq − (x − a)2 (x + a)2 → i → i → i x < −a (Toward the right) −a < x < +a (Toward the left) x > +a (Toward the right) (20.11) 668 20 Electric Fields When x a we can take out a factor of x from each denominator of the brackets of the last formula for x > +a and then expand each of these terms by binomial expansion Therefore, we get: → → 1 − i 2 (x − a) (x + a) a −2 a −2 → kq − 1+ = 1− i x x x 2a kq 2a − ··· − − + ··· = 1+ x x x kq 4a → k (2a q) → i = i x2 x x3 E = kq (20.12) → i x a For x −a, we can find an identical expression but with |x| instead of x in the last formula The product of the positive charge q and the length of the dipole 2a is called the magnitude of the electric dipole moment, p = 2a q The direction of → p is taken → to be from the negative charge to the positive charge of the dipole, i.e → p = pi Using this definition, we have: ⎧ → p ⎪ ⎪ x ⎨2k → x E = → ⎪ ⎪2k p x ⎩ |x|3 a −a → (→ p = 2aq i ) (20.13) Thus, at far distances, the electric field along the x-axis is proportional to the electric dipole moment → p and varies as 1/|x | Electric Field Along the Perpendicular Bisector of a Dipole Axis Let us assume that a point P lies on the y-axis, i.e along the perpendicular bisector of the line joining the dipole charges, see Fig 20.9 Substitute x with in Eq 20.10 to get: → E = a2 kq → rˆ+ − → rˆ− + y2 (20.14) From Fig 20.9, we see that: → → rˆ+ = − cos θ i + → → → sin θ j , → rˆ− = cos θ i + sin θ j , cos θ = a/ a2 + y2 (20.15) 20.3 The Electric Field of an Electric Dipole 669 Fig 20.9 The electric field → → E+ → E = E+ + E− at point P(0, y) y along the y-axis of an electric dipole lying along the x-axis θ with a length 2a E P θ E− y r− r+ r+ r− -q - θ θ +q + x (a ,0 ) ( -a , ) Substituting these relations in Eq 20.14 we get: → E = −k → → 2a q p = −k i (a2 + y2 )3/2 (a2 + y2 )3/2 → (→ p = 2a q i ) (20.16) When |y| a, we can neglect a2 when we compare it with y2 in the denominator bracket and write: ⎧ → p ⎪ ⎪ − k y a ⎪ ⎨ y3 → → (→ (20.17) E = p = 2aq i ) → ⎪ p ⎪ ⎪ ⎩ − k y −a |y| Thus, at far distances, the electric field along the perpendicular bisector of the line joining the dipole charges is proportional to the electric dipole moment → p and varies as 1/|y|3 Generally, this inverse cube dependence at a far distance is a characteristic of a dipole Example 20.3 (The Dipole Field Along the Dipole Axis) A proton and an electron separated by × 10−10 m form an electric dipole, see Fig 20.10 Use exact and approximate formulae to calculate the electric field on the x-axis at a distance 20 × 10−10 m to the right of the dipole’s center 670 20 Electric Fields Electron -e Proton - -a +e E− + P E+ x +a a = 10−10 m 20 × 10−10 m Fig 20.10 Solution: In this problem we have a = 10−10 m, q = e = 1.6 × 10−19 C, x = 20 × 10−10 m, ke = (9 × 109 N.m2 /C2 ) (1.6 × 10−19 C) = 1.44 × 10−9 N.m2 /C, x − a = 19 × 10−10 m, and x + a = 21 × 10−10 m Using the exact formula given by Eq 20.11 in the case of x > +a, we have: E = ke 1 − (x − a)2 (x + a)2 = (1.44 × 10−9 N.m2 /C) 1 − (19 × 10−10 m)2 (21 × 10−10 m)2 = (1.44 × 10−9 N.m2 /C)[2.770 × 1017 m−2 − 2.268 × 1017 m−2 ] = 7.236 × 107 N/C On the other hand, we have x a and we can use the approximate formula given by Eq 20.13 as follows: 2a e 4a p = 2k = ke 3 x x x (4 × 10−10 m) = (1.44 × 10−9 N.m2 /C) (20 × 10−10 m)3 E = 2k = 7.200 × 107 N/C Clearly this calculation is a good approximation when x/a = 20 20.4 Electric Field of a Continuous Charge Distribution The electric field at point P due to a continuous charge distribution shown in Fig 20.11 can be evaluated by: (1) Dividing the charge distribution into small elements, each of charge located relative to point P by the position vector → r =r → rˆ n n n qn that is 20.4 Electric Field of a Continuous Charge Distribution → En due to the nth element as follows: (2) Using Eq 20.4 to evaluate the electric field → En = k 671 qn → rˆn rn2 (20.18) (3) Evaluating the order of the total electric field at P due to the charge distribution by the vector sum of all the charge elements as follows: → qn → rˆn rn2 E ≈k n (20.19) (4) Evaluating the total electric field at P due to the continuous charge distribution in the limit qn → as follows: → E = k lim qn →0 n qn → rˆn = k rn2 dq → rˆ r2 (20.20) where the integration is done over the entire charge distribution Fig 20.11 The electric field P → E at point P due to a continuous charge distribution qn is the vector sum of all the fields → En rn En (n = 1, 2, · · · ) due to the charge elements qn (n = 1, 2, · · · ) of the charge distribution Now we consider cases were the total charge is uniformly distributed on a line, on a surface, or throughout a volume It is convenient to introduce the charge density as follows: (1) When the charge Q is uniformly distributed along a line of length L, the linear charge density λ is defined as: λ= Q L (20.21) where λ has the units of coulomb per meter (C/m) (2) When the charge Q is uniformly distributed on a surface of area A, the surface charge density σ is defined as: 672 20 Electric Fields σ = Q A (20.22) where σ has the units of coulomb per square meter (C/m2 ) (3) When the charge Q is uniformly distributed throughout a volume V, the volume charge density ρ is defined as: ρ= Q V (20.23) where ρ has the units of coulomb per cubic meter (C/m3 ) Accordingly, the charge dq of a small length dL, a small surface of area dA, or a small volume dV is respectively given by: dq = λ dL, dq = σ dA, dq = ρ dV (20.24) 20.4.1 The Electric Field Due to a Charged Rod For a Point on the Extension of the Rod Figure 20.12 shows a rod of length L with a uniform positive charge density λ and total charge Q In this figure, the rod lies along the x-axis and point P is taken to be at the origin of this axis, located at a constant distance a from the left end When we consider a segment dx on the rod, the charge on this segment will be dq = λ dx y P dx x E + + + + + + + + + + + + dq a x L → Fig 20.12 The electric field E at point P due to a uniformly charged rod lying along the x-axis The magnitude of the field due to a segment of charge dq at a distance x from P is k dq/x The total field is the vector sum of all the segments of the rod → The electric field dE at P due to this segment is in the negative x direction and has a magnitude given by: