Electric Potential PDF

B. Sc. Physics (H.R.K) Chapter 30: Electric Potential ELECTRIC POTENTIAL 30.1 Potential Difference Potential difference ‘∆ ’ between two point is defined as “the amount of work done ‘∆ ’ per unit charge ‘...

B. Sc. Physics (H.R.K) Chapter 30: Electric Potential ELECTRIC POTENTIAL 30.1 Potential Difference Potential difference ‘∆ ’ between two point is defined as “the amount of work done ‘∆ ’ per unit charge ‘ ’ in moving it from one point to the other against the electric field and ∆ by keeping the system in equilibrium”. Mathematically ∆ Suppose a unit positive test charge ‘ ’ is moved from one point ‘ ’ to the point ‘ ’ in the electric field ‘ ’ of a large positive charge ‘ ’ as shown in figure below, The work done in moving ‘ ’ from point ‘ ’ to the point ‘ ’ against the electric field ‘ ’ is →. The electrical force of magnitude ‘ ’ must have to supplied in order to move ‘ ’ against the electric field. Therefore →. →. →.. → ∵ → ∆ Therefore, the electrical potential difference between two points in an electrical field will be. ---------------------- (1) 30.2 Absolute Electrical Potential at a Point Absolute electric potential at a point is defined as “the amount of work done per unit charge in moving it from infinity to a specific field point against the electric field and by keeping the system in equilibrium”. 33 Written and composed by: Prof. Muhammad Ali Malik, Govt. Degree College, Naushera, 03016775811 B. Sc. Physics (H.R.K) Chapter 30: Electric Potential To find the absolute potential, the reference point is selected at which potential is zero. ∞ 0 This point is situated at infinity i.e., out of the electric field. Thus, in equation (1) ⟹ 0. ∞ If the distance from the point ‘ ’ to the charge ‘ ’ is ‘ ’, then in general !. " 30.3 Expression for the Electric Potential Difference due to a Point Charge The potential difference between two points is the amount of work done per unit charge ‘ ’ in moving it from one point to the other against the electric field ‘#’. Mathematically, it is described as. But ̂ $ %&' !( ̂. $ So, %&' !( ⟹ ! !̂. %&' ! !( ∴E is directed radially outward, therefore ̂ // ⟹ ! %&' ! !( $ ! $ ! ⟹ , , , , %&' ! ! %&' ! ! ⟹ -. $ $ %&' ! ! ---------------------- (2) This is the expression for the potential difference between two points ‘ ’ and ‘ ’. 30.4 Expression for the Absolute Electric Potential due to a Point Charge The electric potential at any point is the amount of work done per unit charge in moving a unit positive charge (test charge) from infinity to that point, against the electric field. ∞ 0; ∞ If the point ‘ ’ is at infinity then 34 Written and composed by: Prof. Muhammad Ali Malik, Govt. Degree College, Naushera, 03016775811 B. Sc. Physics (H.R.K) Chapter 30: Electric Potential ⟹ 0 -. 0 Putting this value in equation (2), we get $ $ 1 ∴ %&' ! " ∞ ⟹ -. $ %&' ! $ In general, the electric potential at point due to a point charge ‘ ’ is %&' ! 30.5 Electric Potential Due to a Collection of Point Charges Let “ $ , 1, 2,.......... , 3 ” are the ‘n’ point charges which are at distances “ $ , 1 , 2 ,.......... , 3 ” from a point ‘4’, as shown in the figure below Now if $, 1, 2,........ , 3 be the electric potential at a field point ‘4’ due to the point charges $, 1, 2,.......... , 3 respectively. Then, the total electric potential at a field point ‘4’ due to assembly of ‘n’ point charges will be; $ 5 1 5 2 5........ 5 3 ---------------------- (3) $ where 6 $ %&' !6 = Electric Potential at a Field Point ‘4’ due to Point Charge ‘ $ ’ = $ ( 1 %&' !( = Electric Potential at a Field Point ‘4’ due to Point Charge ‘ 1 ’ = $ 7 2 %&' !7 = Electric Potential at a Field Point ‘4’ due to Point Charge ‘ 2 ’ = ⋮ ⋮ ⋮ ⋮ ⋮ ⋮ ⋮ ⋮ ⋮ ⋮ ⋮ ⋮ ⋮ ⋮ $ 9 3 3’ %&' !9 = Electric Potential at a Field Point ‘4’ due to Point Charge ‘ = 5 5 5........ 5 $ $ $ $ Putting values in equation (3), we get 6 ( 7 9 %&' !6 %&' !( %&' !7 %&' !9 : 5 5 5........ 5 ; $ 6 ( 7 9 %&' !6 !( !7 !9 = ∑3>?$ $ = %&' != = ---------------------- (4) This is the net potential at any point ‘4’ due to collection of ‘n’ point charges. 35 Written and composed by: Prof. Muhammad Ali Malik, Govt. Degree College, Naushera, 03016775811 B. Sc. Physics (H.R.K) Chapter 30: Electric Potential 30.6 Electric Potential due to a Dipole Consider two point charges 5 and – of equal magnitude lying distance ‘A’ apart as shown in the figure below. The electric potential at point ‘4’ is the sum of potential due to ‘5 ’ and ‘ ’. If B and _ are the electric potential due to the charges 5 and –. Therefore, B5 _ 1 5 1 5 4EF $ 4EF 1 1 1 G H 4EF $ 1 G H 1 $ 4EF $ 1 ≫A Normally for a dipole, $ 1 J 1 Therefore, And from figure, 1 $ A KLMN A KLMN G H 4EF 1 1 A KLMN 4EF 1 ∴O A = Dipole Moment 1 O KLMN 4EF 1 1 O KLMN 4EF 2 ∴ PQRSTORUTVW VA ATXTATVW U 1 Y. 4EF 2 This is the expression of electric potential at any point ‘4’ due to a dipole. 30.7 Electric Potential of Continuous Charge Distribution 30.7.1 Electric Potential due to a Ring of Charge Consider a ring of positive charge of radius ‘Z’ as shown in the figure.We want to find out the electric potential at point ‘4’ which is at the distance ‘[’ from the plane of ring. so it has constant linear charge density \. For an As the charge is distributed uniformly over it, A infinitesimal length element ‘AM’ of ring, \ AM A \ AM The electric potential due to the charge A at point ‘4’ is given by; 36 Written and composed by: Prof. Muhammad Ali Malik, Govt. Degree College, Naushera, 03016775811 B. Sc. Physics (H.R.K) Chapter 30: Electric Potential A $ ] $ ^ ]_ %&' ! %&' ! A $ ^ ]_ %&' √a ( B b( ∴ c Ld cTWQ e, √[ 1 5 Z 1 1 \ AM The total value of electric potential is A 4EF √[ 1 5 Z 1 1 1 \ AM 4EF √[ 1 5 Z 1 ∴ \ AM 1 Therefore, 4EF √[ 1 5 Z 1 This is the expression of electric potential at any point ‘4’ due to a ring of charge. 30.7.2 Electric Potential due to a Disk of Charge Consider a circular disk of uniform surface charge density as shown in the figure below. We want to find out the electric potential at point ‘4’ which is at the distance ‘[’ from the plane of disk. Consider a small element of the disk in the ring shape of radius ‘f’ and width ‘Af’. If ‘A ’ is the A charge on this element of ring, then g Ah ∴ ije e Ah TM Sje TM Sje e Lk ReVWSj eRedeVS A g Ah A g 2Ef Af ∴ Ah 2Ef Af We know the electric potential due to the ring of charge which has the radius f is given by; 1 A A 4EF 1 g 2Ef Af A 4EF [ 1 5 f 1 The electrical potential at point ‘4’ due to whole disk is 1 m?b g 2Ef Af 4EF m? [1 5 f1 g m?b $ [ 1 5 f1 n 1 2f Af 4F m? 1 Z [2 5 f2 51 g 2 o o 4F 1 2 0 1 g 1 p [2 5 Z 2 [ 2q 2 2 2F 37 Written and composed by: Prof. Muhammad Ali Malik, Govt. Degree College, Naushera, 03016775811 B. Sc. Physics (H.R.K) Chapter 30: Electric Potential g r 2 p [ 5 Z –[ q 2 2F This is the expression of electric potential at point ‘4’ due to disk of charge. Special Case: If [ ≫ Z, then 6 √[ 1 5 Z1 = [ 1 5 Z1 ( 6 b( ( = [ -1 5. a( $ b( [ -1 5 5 ………. 1 a( b( ⇒ √[ 1 5 Z1 -[ 5. ∴ ueWReKSTVW vTWje we dM 1a Therefore, g Z 2 gZ 2 p[ 5 –[ q 2F 2[ 4F [ 2 g EZ g EZ 2 4EF [ 4EF [ 2 ∴ g EZ SLS R Kj We LV ATMx) 1 4EF [ Thus, the disk of charge behave like a point charge for the case [ ≫ Z. 30.8 Equipotential Surfaces If all the points on a surface have the same value of electric potential, then, it is known as equipotential surface. The examples of some equipotential surfaces are given below In case of uniform electric field, the equipotential surfaces are the planes as shown by the dashed line in fig. (a). The potential at points y$ , y1 and y2 is same. So, no work will be done in moving a test charge from y$ to y1 or y2. The concentric spheres about a point charge ‘5 ’ are the equipotential surfaces as shown in the fig. (b) by dashed lines. The equipotential surfaces of a dipole are shown by the dashed lines in fig. (c) Fig. (a) Fig. (b) Fig. (c) 38 Written and composed by: Prof. Muhammad Ali Malik, Govt. Degree College, Naushera, 03016775811 B. Sc. Physics (H.R.K) Chapter 30: Electric Potential 30.9 Calculating Electric Field from the Electric Potential Consider a test charge ‘ ’ is displaced in an electric field form point ‘4’ to point ‘€’. If the potential difference between the points ‘4’ and ‘€’ is A , then the work done ‘A ’ by the electric field is A A ---------------------- (1) If the test charge is placed in an electric field, then the electrical force on the test charge is So, the corresponding work done is A. d. d ---------------------- (2) Comparing eq. (1) and (2). d A # AM KLMN A ]‚ # KLMN ]_ Consider ‘# KLMN #_ ’ which is the component of ‘ ’ along ‘A ’ ]‚ #_ ]_ There is only one direction in which the rate of change of electric potential w r t position is maximum, which is direction of electric field. Thus ]‚ #_ : ]_ ; ƒ „ ]‚ ]_ The maximum value of at a given point is called the ‘potential gradient’. For any arbitrary direction of ‘A ’ …‚ …‚ …‚ #„ ; #† ; #a …„ …† …a Therefore, #„ ‡̂ +#† ˆ̂ 5 #a x‰ x‰ …‚ …‚ …‚ ‡̂ + ˆ̂ 5 …„ …† …a x‰ … … … ‡̂ + ˆ̂ 5 …„ …† …a Š grad This expression states that electric field is negative gradient of electric potential. 39 Written and composed by: Prof. Muhammad Ali Malik, Govt. Degree College, Naushera, 03016775811 B. Sc. Physics (H.R.K) Chapter 30: Electric Potential Question: Find the expression of electric field from electric potential due to a point charge. 1 Solution: the electric potential due to a point charge is 4EF $ = ∴ Œ 1 + U 1 + [ 1 %&' Œ„ ( B † ( Ba ( grad =( ‡̂ + ˆ̂ + x‰) … … … …„ …† …a =( ‡̂ + ˆ̂ + x‰) ---------------------- (3) …‚ …‚ …‚ …„ …† …a Now consider 1 = Ž 4EF …‚ … …„ …„ 0 r 2 + U2 +[2 1 − 2 (2 ) = 4EF( 1 + U 1 + [ 1 )2⁄1 ‘ = − = − ‘ 4EF ( + U + [ 1 )2⁄1 1 1 4EF 2 ‘ U U Similarly = − = − ‘U 4EF ( + U + [ 1 )2⁄1 1 1 4EF 2 ‘ [ [ = − = − ‘[ 4EF ( 1 + U 1 + [ 1 )2⁄1 4EF 2 ‰ ‡̂ + Uˆ̂ + [x Putting these values in eq. (3) grad =− ’ ” 4EF0 3 ∴ = ‡̂ + Uˆ̂ + [x‰ grad =− - 3. 4EF0 ̂ grad =− G H 4EF0 3 ∴ = ̂ ̂ grad =− G 2H 4EF0 !̂ −grad = - 2. ∴ = − grad 4EF0 = Therefore - (. %&' ! which is the expression of electric field intensity due to a point charge. 30.10 Electric Potential and Electric Field Inside and Outside an Isolated Conductor If the excess of charge is placed on an isolated conductor, then it moves entirely on to the outer surface of the conductor. In equilibrium, none of the charge is inside the body of conductor or on any interior surface (even the conductor has internal cavities). This property of the conductors can be stated in terms of electrical potential as 40 Written and composed by: Prof. Muhammad Ali Malik, Govt. Degree College, Naushera, 03016775811 B. Sc. Physics (H.R.K) Chapter 30: Electric Potential “An excess of charge placed on an isolated conductor distributes itself on the surface so that all points of conductor come to the same potential”. Now consider an isolated conductor in the form of a spherical shell having uniform surface charge density. In this case, the electric field is radial and its value is maximum on the surface of the spherical shell. Also, it decreased as we move away from the sphere because #∝ $ !( The electric field at any point inside sphere is zero. The electric potential decreases as we move away from the sphere, because ∝ $ ! The value of electric potential is maximum at the surface of sphere and it remains constant at every point inside the sphere. This is so because if a test charge is pushed inside the sphere through the hole, then it experiences no force and no work will be done. Hence the electric potential at all the points inside the sphere is same. As the electric potential inside and on the surface of sphere has the same constant value, its derivative will be zero. hM = —LVMS VS ]‚ ] ∴# KLVMS VS 0 ]_ ]_ So the electric field inside the sphere is zero and it has maximum value at the surface of sphere. The variation of electric potential with respect to ‘ ’ and variation electric field with respect to ‘ 1 ’ is shown in the figure below. 41 Written and composed by: Prof. Muhammad Ali Malik, Govt. Degree College, Naushera, 03016775811