Electric Potential: Continuous Charge Distribution - Physics Lecture Notes - PDF

General Physics II (4103) Electricity Lec. 07 5. Continuous Charge Distribution If the charge distribution is continuous, the potential at a point 𝑷 can be found by summing over the contributions from individual differential elements of charge 𝒅𝒒....

General Physics II (4103) Electricity Lec. 07 5. Continuous Charge Distribution If the charge distribution is continuous, the potential at a point 𝑷 can be found by summing over the contributions from individual differential elements of charge 𝒅𝒒. Consider the charge distribution shown in Fig. (6). Taking infinity as our reference point with zero potential, the electric potential at 𝑷 due to 𝒅𝒒 is 𝟏 𝒅𝒒 𝒅𝑽 = (𝟏𝟓) 𝟒𝝅𝝐𝒐 𝒓 Summing over contributions from all differential elements, we have 𝟏 𝒅𝒒 𝑽 = ∫ 𝒅𝑽 = ∫ (𝟏𝟔) 𝟒𝝅𝝐𝒐 𝒓 Figure 6. Continuous charge distribution. Electric Potential on The Axis of a Charged Ring: Figure (7) shows a uniformly charged ring of radius 𝒂 and charge 𝑸 in the 𝒛 = 𝟎 plane and centered at the origin. The distance from an element of charge 𝒅𝒒 to the field point 𝑷 on the axis of the ring is 𝒓 = √𝒛𝟐 + 𝒂𝟐 Because this distance is the same for all elements of charge on the ring, we can remove this term from the integral in Eqn. (16). The potential at point due to the ring is thus 𝒌 𝒅𝒒 𝒌 𝒌𝑸 𝑽=∫ = ∫ 𝒅𝒒 = 𝒓 𝒓 𝒓 1 General Physics II (4103) Electricity Lec. 07 Or 𝒌𝑸 𝑽= (𝟏𝟕) √𝒛𝟐 + 𝒂𝟐 Note that when |𝒛| is much greater than 𝒂 the potential approaches 𝒌𝑸⁄|𝒛| the same as the potential due to a point charge 𝑸 at the origin. Figure (7) Electric potential 𝑽 for a Charged Disk Find the potential on the axis of a disk of radius 𝑹 that carries a total charge 𝑸 distributed uniformly on its surface. Solution: We take the axis of the disk to be the 𝒛 axis, and we treat the disk as a set of ring charges. The ring of radius 𝒂 and thickness 𝒅𝒂 in Fig. (8) has an area of 𝒅𝑨 = 𝟐𝝅𝒂 𝒅𝒂 Figure (8) 2 General Physics II (4103) Electricity Lec. 07 The charge of the ring is 𝒅𝒒 = 𝝈𝒅𝑨 = 𝝈𝟐𝝅𝒂 𝒅𝒂 Where; 𝑸 𝝈= 𝝅𝑹𝟐 It is the surface charge density. The potential at point 𝑷 due to the charge on this ring is given by 𝒌𝒅𝒒 𝒅𝑽 = √𝒛𝟐 + 𝒂𝟐 We then integrate from 𝒂 = 𝟎 to 𝒂 = 𝑹 to find the total potential due to the charge on the disk. 1. Write the potential 𝒅𝑽 at point due to the charged ring of radius 𝒂: 𝒌𝒅𝒒 𝒌𝝈𝟐𝝅𝒂 𝒅𝒂 𝒅𝑽 = = √𝒛𝟐 + 𝒂𝟐 √𝒛𝟐 + 𝒂𝟐 2. Integrate from 𝒂 = 𝟎 to 𝒂 = 𝑹: 𝑹 𝑹 𝒌𝝈𝟐𝝅𝒂 𝒅𝒂 𝟐𝒂 𝒅𝒂 𝑽=∫ = 𝒌𝝈𝝅 ∫ 𝟎 √𝒛𝟐 + 𝒂𝟐 𝟎 √𝒛𝟐 + 𝒂𝟐 3. The integral is of the form 𝟏 ∫ 𝒖𝒏 𝒅𝒖 With 𝒖 = 𝒛𝟐 + 𝒂𝟐 , 𝒅𝒖 = 𝟐𝒛𝒅𝒛 and 𝒏 = − 𝟐 When 𝒂 = 𝟎, 𝒖 = 𝒛𝟐 and when 𝒂 = 𝑹, 𝒖 = 𝒛𝟐 + 𝑹𝟐 𝒛𝟐 +𝑹𝟐 𝒛𝟐 +𝑹𝟐 𝒖𝟏⁄𝟐 𝑽 = 𝒌𝝈𝝅 ∫ 𝒖−𝟏⁄𝟐 𝒅𝒖 = 𝒌𝝈𝝅 [ ] 𝒛𝟐 𝟏⁄𝟐 𝒛𝟐 𝑽(𝒛) = 𝟐𝒌𝝈𝝅 [√𝒛𝟐 + 𝑹𝟐 − √𝒛𝟐 ] (𝟏𝟖) 4. Rearranging this result to find 𝑽 gives 𝑹𝟐 𝑽(𝒛) = 𝟐𝒌𝝈𝝅|𝒛| [√𝟏 + − 𝟏] (𝟏𝟗) 𝒛𝟐 Then, the potential on the axis of a uniformly charged disk in the plane is 3 General Physics II (4103) Electricity Lec. 07 𝑹𝟐 𝑽(𝒛) = 𝟐𝒌𝝈𝝅|𝒛| [√𝟏 + − 𝟏] (𝟐𝟎) 𝒛𝟐 For |𝒛| ≫ 𝑹 the potential function 𝑽 should approach the potential function of a point charge 𝑸 at the origin. That is, we expect that for large |𝒛|, 𝒌𝑸 𝑽= |𝒛| To approximate our result for |𝒛| ≫ 𝑹, we use the binomial expansion: 𝟏⁄𝟐 𝑹𝟐 𝟏 𝑹𝟐 (𝟏 + 𝟐 ) =𝟏+ + ⋯+ ⋯ 𝒛 𝟐 𝒛𝟐 Then 𝟏 𝑹𝟐 𝟏 𝑹𝟐 𝑽(𝒛) = 𝟐𝒌𝝈𝝅|𝒛| [(𝟏 + + ⋯ + ⋯ ) − 𝟏] = 𝟐𝒌𝝈𝝅|𝒛| 𝟐 𝒛𝟐 𝟐 𝒛𝟐 𝒌(𝝈𝝅𝑹𝟐 ) 𝒌𝑸 𝑽(𝒛) ≈ ≈ (𝟐𝟏) |𝒛| |𝒛| 6. Deriving Electric Field from the Electric Potential In Eqn. (1) we established the relation between ⃗𝑬 and 𝑽. If we consider two points which are separated by a small distance 𝒅𝒔 ⃗ , the following differential form is obtained: ⃗ ∙ 𝒅𝒔 𝒅𝑽 = −𝑬 ⃗ (𝟏𝟕) In Cartesian coordinates, ̂ ⃗⃗ = 𝑬𝒙 𝒊̂ + 𝑬𝒚 𝒋̂ + 𝑬𝒛 𝒌 𝑬 and ̂ ⃗ = 𝒅𝒙 𝒊̂ + 𝒅𝒚 𝒋̂ + 𝒅𝒛 𝒌 𝒅𝒔 we have ̂ ) ∙ (𝒅𝒙 𝒊̂ + 𝒅𝒚 𝒋̂ + 𝒅𝒛 𝒌 𝒅𝑽 = −(𝑬𝒙 𝒊̂ + 𝑬𝒚 𝒋̂ + 𝑬𝒛 𝒌 ̂) ∴ 𝒅𝑽 = 𝑬𝒙 𝒅𝒙 + 𝑬𝒚 𝒅𝒚 + 𝑬𝒛 𝒅𝒛 (𝟏𝟖) But’ ∵ 𝑽 = 𝑽(𝒙, 𝒚, 𝒛) 4 General Physics II (4103) Electricity Lec. 07 𝝏𝑽 𝝏𝑽 𝝏𝑽 ∴ 𝒅𝑽 = 𝒅𝒙 + 𝒅𝒚 + 𝒅𝒛 (𝟏𝟗) 𝝏𝒙 𝝏𝒚 𝝏𝒛 which implies 𝝏𝑽 𝝏𝑽 𝝏𝑽 𝑬𝒙 = − , 𝑬𝒚 = − , 𝑬𝒛 = − (𝟐𝟎) 𝝏𝒙 𝝏𝒚 𝝏𝒛 Then, the electric field can be written as 𝝏𝑽 𝝏𝑽 𝝏𝑽 ⃗𝑬 ̂ = −( ⃗ = 𝑬𝒙 𝒊̂ + 𝑬𝒚 𝒋̂ + 𝑬𝒛 𝒌 𝒊̂ + 𝒋̂ + ̂) 𝒌 (𝟐𝟏) 𝝏𝒙 𝝏𝒚 𝝏𝒛 Mathematically, we can think of ⃗𝑬 as the negative of the gradient of the electric potential 𝑽. Physically, the negative sign implies that if 𝑽 increases as a positive charge moves along some direction, say 𝒙, with , then there is a non-vanishing component of 𝑬 ⃗⃗ in the opposite direction (−𝑬𝒙 ≠ 𝟎). If the charge distribution possesses spherical symmetry, then the resulting electric field is a function of the radial distance 𝒓, i.e., ⃗𝑬 = 𝑬𝒓 𝒓̂ In this case, 𝒅𝑽 = −𝑬𝒓 𝒅𝒓 If 𝑽(𝒓) is known, then ⃗𝑬 may be obtained as 𝒅𝑽 ⃗𝑬 ⃗ = 𝑬𝒓 𝒓̂ = − ( ) 𝒓̂ (𝟐𝟐) 𝒅𝒓 For example, the electric potential due to a point charge 𝒒 is 𝟏 𝒒 𝑽(𝒓) = 𝟒𝝅𝝐𝒐 𝒓 Using the above formula, the electric field is simply 𝒅𝑽 𝒒 𝒅 𝟏 𝒒 𝟏 ⃗⃗ = − ( ) 𝒓̂ = − 𝑬 ( ) 𝒓̂ = − (− 𝟐 ) 𝒓̂ 𝒅𝒓 𝟒𝝅𝝐𝒐 𝒅𝒓 𝒓 𝟒𝝅𝝐𝒐 𝒓 𝟏 𝒒 ∴ ⃗⃗ (𝒓) = 𝑬 𝒓̂ 𝟒𝝅𝝐𝒐 𝒓𝟐 5 General Physics II (4103) Electricity Lec. 07 ⃗⃗ for a Potential that Varies with 𝒙 Example (1). 𝑬 Find the electric field for the electric potential function given by 𝑽(𝒙) = 𝟏𝟎𝟎 𝑽 − (𝟐𝟓 𝑽⁄𝒎)𝒙 Solution This potential function depends only on 𝒙. Use 𝒅𝑽(𝒙) 𝑬𝒙 = − 𝒅𝒙 And solve for 𝑬𝒙. Because the potential does not vary with 𝒚 or 𝒛; 𝑬𝒚 = 𝑬𝒛 = 𝟎 𝒅𝑽(𝒙) 𝒅 𝑬𝒙 = − =− [𝟏𝟎𝟎 𝑽 − (𝟐𝟓 𝑽⁄𝒎)𝒙] = −[−(𝟐𝟓 𝑽⁄𝒎)] = (𝟐𝟓 𝑽⁄𝒎) 𝒅𝒙 𝒅𝒙 𝑽 𝑬𝒙 = 𝟐𝟓 𝒎 Example (2) Potential and field of a ring of charge We found that for a ring of charge with radius 𝒂 and total charge 𝑸. the potential at a point 𝑷 on the ring axis a distance 𝒙 from the center is 𝟏 𝑸 𝑽(𝒙) = 𝟒𝝅𝝐𝒐 √𝒙𝟐 + 𝒂𝟐 Find the electric field at 𝑷. Solution: From the symmetry of the charge distribution shown in Fig. (7), the electric field along the symmetry axis of the ring can have only an x-component. 6 General Physics II (4103) Electricity Lec. 07 We find it using the first of Eqns. (21). The x-component of the electric field is 𝝏𝑽 𝝏 𝟏 𝑸 𝑸 𝝏 𝟏 𝑬𝒙 = − =− [ ]=− [ ] 𝝏𝒙 𝝏𝒙 𝟒𝝅𝝐𝒐 √𝒙𝟐 + 𝒂𝟐 𝟒𝝅𝝐𝒐 𝝏𝒙 √𝒙𝟐 + 𝒂𝟐 𝑸 𝟏 𝑬𝒙 = − [− (𝒙𝟐 + 𝒂𝟐 )−𝟑⁄𝟐 (𝟐𝒙)] 𝟒𝝅𝝐𝒐 𝟐 𝟏 𝑸𝒙 ∴ 𝑬𝒙 (𝒙) = 𝟒𝝅𝝐𝒐 (𝒙 + 𝒂𝟐 )𝟑⁄𝟐 𝟐 Example 3. Calculating Electric Field from Electric Potential Suppose the electric potential due to a certain charge distribution can be written in Cartesian Coordinates as 𝑽(𝒙, 𝒚, 𝒛) = 𝑨𝒙𝟐 𝒚𝟐 + 𝑩𝒙𝒚𝒛 where 𝑨 and 𝑩 are constants. What is the associated electric field? Solution: The electric field can be found by using Eqn. (21): 𝝏𝑽 𝝏 𝑬𝒙 = − =− (𝑨𝒙𝟐 𝒚𝟐 + 𝑩𝒙𝒚𝒛) = −(𝟐𝑨𝒙𝒚𝟐 + 𝑩𝒚𝒛) 𝝏𝒙 𝝏𝒙 𝝏𝑽 𝝏 𝑬𝒚 = − =− (𝑨𝒙𝟐 𝒚𝟐 + 𝑩𝒙𝒚𝒛) = −(𝟐𝑨𝒙𝟐 𝒚 + 𝑩𝒙𝒛) 𝝏𝒚 𝝏𝒚 𝝏𝑽 𝝏 𝑬𝒛 = − = − (𝑨𝒙𝟐 𝒚𝟐 + 𝑩𝒙𝒚𝒛) = −(𝟎 + 𝑩𝒙𝒚) = −𝑩𝒙𝒚 𝝏𝒛 𝝏𝒛 Therefore, the electric field is ̂ ⃗𝑬(𝒙, 𝒚, 𝒛) = (−𝟐𝑨𝒙𝒚𝟐 − 𝑩𝒚𝒛)𝒊̂ − (𝟐𝑨𝒙𝟐 𝒚 + 𝑩𝒙𝒛)𝒋̂ − (𝑩𝒙𝒚)𝒌 7 General Physics II (4103) Electricity Lec. 07 7. Equipotential Surfaces and Field Lines Suppose a system in two dimensions has an electric potential 𝑽(𝒙, 𝒚). The curves characterized by constant 𝑽(𝒙, 𝒚) are called equipotential curves. Examples of equipotential curves are depicted in Fig. (8) below. Figure 8. Equipotential curves. In three dimensions we have equipotential surfaces and they are described by 𝑽(𝒙, 𝒚, 𝒛) = 𝒄𝒐𝒏𝒔𝒕𝒂𝒏𝒕 The potential 𝑽 has the same value everywhere on an equipotential surface. If a test charge on an equipotential surface is given a small displacement 𝒅𝒍 parallel to the surface, ⃗ ∙ 𝒅𝒍 = 𝟎 𝒅𝑽 = −𝑬 ⃗⃗ ∙ 𝒅𝒍 is zero for any 𝒅𝒍 parallel to the surface, 𝑬 Because 𝑬 ⃗ must either be zero or be perpendicular to any and every 𝒅𝒍 that is parallel to the surface. The only way 𝑬⃗⃗ can be perpendicular to every 𝒅𝒍 parallel to the surface is for 𝑬 ⃗ to be normal to the surface. Therefore, we conclude that electric field lines are normal to any equipotential surfaces they intersect. Figs. (9 & 10) shows two arrangements of charges. The field lines in the plane of the charges are represented by red lines, and the intersections of the equipotential surfaces with this plane (that is, cross sections of these surfaces) are shown as blue lines. The actual equipotential surfaces are three-dimensional. At each crossing of an equipotential and a field line, the two are perpendicular. 8 General Physics II (4103) Electricity Lec. 07 Figure (9) Figure (10) Figs. (11) and (12) show equipotential surfaces near a spherical conductor and a non-spherical conductor. 9 General Physics II (4103) Electricity Lec. 07 Figure 11. Equipotential surfaces and electric field lines outside a uniformly charged spherical conductor. Figure 12. Equipotential surfaces and electric field lines outside a non-spherical conductor. The properties of equipotential surfaces can be summarized as follows: (i) The electric field lines are perpendicular to the equipotential and point from higher to lower potentials. (ii) By symmetry, the equipotential surfaces produced by a point charge form a family of concentric spheres, and for constant electric field, a family of planes perpendicular to the field lines. (iii) The tangential component of the electric field along the equipotential surface is zero, otherwise non-vanishing work would be done to move a charge from one point on the surface to the other. (iv) No work is required to move a particle along an equipotential surface. 10 General Physics II (4103) Electricity Lec. 07 11