electric field intensity

1. CHAPTER (2) Electric Charges, Electric Charge Densities and Electric Field Intensity

2. Charge Configuration a) Point Charge: The concept of the point charge is used when the dimensions of an electric charge distribution are very small compared to the distance to the neighboring charges, i.e. the point charge is occupying a very small physical space b) Distributed Charge The charge may be distributed along a line, among a surface or among a volume. (1) Line Charge: The line charge density ?? is defined as the charge per unit length. + + + + + + + + dl + + dQ = ρL dl + + + + + + ∆? ∆?�?? � � ??= ???∆?→? ??= ?? ??�?? � � So, ?? = ???? The total charge Q of the line can be determined by

3. ? ? [?] ? = � ?? = � ???? ? ? For uniform line charge, where ?? is constant ? = ??? [?] U(2) Surface Charge: The surface charge density ?? is defined as the charge per unit surface area. ∆? ∆?�? ??= ???∆?→? � � ?? ??= ?? ??�? � � ?? So, ?? = ???? The total charge Q of the surface can be determined by

4. ? = �?? = �???? For uniform surface charge, where ?? is constant ? = ??? [?] U3) Volume charge density (??) ∆? ∆?�? ??= ???∆?→? � � ?? ??�? ??= ?? dv � � ?? dQ = ρv dv So, ?? = ???? The total charge Q of the surface can be determined by ? = �?? = �???? UFor uniform volume chargeU, where ?? is constant ? = ??? [?]

5. Example: A uniform spherical volume charge density distribution contains a total charge of 10-8 C, if the radius of the sphere =2x10-2 m. Find ρ ρv. Solution: π π 4 4 − − − − 8 2 3 2 3 6 2 = = = = = 10 , 2 * 10 , 2 ( * 10 ) 8 * 10 Q C r m V r m 3 3 . − 8 Q 10 − 4 − ρ ∴ = = = 3 . 2 98 * 10 C m v π 4 V − 8 * 8 * 10 3 UExample: A non-uniform spherical volume charge density distribution with ??= ?? charge contained in the volume of the sphere of radius a [m] USolution: ?? = ???? ?? C.mP-3 P. Find the total ? = �?? = �???? = �?? ????????? ?? ??

6. ? ? = ??�??? ??[−????][−????]? ? [?]? ?� ? ? = ??????? Coloumb’s Law: Force Between Two Point Charges: The force between two stationary point charges Q1,Q2is proportional to the product of the two charges and inversely proportional to the square of the distance R between them. ? ��⃗??= ??????? �? ��⃗?? ? ��⃗??= ??????? �? ��⃗?? F F R ˆ a 12 21 12 Q Q 1 2 ???? ???? Where: ? �? ��⃗??= −? �? ��⃗?? unit vector from charge QR1 charge. R to QR2 R

7. and, ??= ??−? Force on Point Charge due to n Point Charges: Force ? ��⃗? on point charge ?? due to n point charges can be determined as ???? ??????? ?=? UExample: Find the force ? ��⃗in vacuum on a point QR1 point charge QR2 (0,1,2)m and QR2 USolution: Q Q a R ˆ ) 4 1 ( ˆ ) 2 0 ( 21 − − − = 22 9 9 4 21 = + + = ∴R ???= ?.?????−???/? R1 t Qt Q 1 ? R2 t ? ��⃗?= � [?] ?? �? ��⃗?? Rn t Q n R=10P-4 P C due to a R=2*10P-5 Rat (2,4,5) P C where QR1 R is centered at point Q Q ∧ 1 2 = = 1 2 F R 21 21 2 πε 3 21 − πε 4 R 4 21 + 0 0 = − − + ˆ ) 5 2 ( R x 3 y z x ˆ y ˆ ˆ 3 2 z

8. − − − ˆ x ˆ y ˆ 3 R 2 3 z = = ˆ a 21 21 R 22 ) ⋅ 21 − − ∗ ∗ ˆ 2 − − − 4 5 ˆ 3 ) ˆ 3 z 0 ( 2 10 ( x y ∴ = F N − π . 8 ∗ ∗ ⋅ 12 2 4 85 10 ( 22 ) 22 Electric Field Intensity at a Point due to Point Charge Q: It is a vector force acting on a unit (+ve) charge. The electric field intensity due to a point located at distance R from the charge Q is given by: ? ∗? ??????? �? ��⃗= Electric Field Intensity at a Point due to Point Charges Q1, Q2,….., Qn: If we have a system of charges Q1, Q2…Qn. the total electric field at a point is the vector sum of all fields due to the different charges. ? ��⃗?= ∑ ?=? [?/?] ?? ?????? ?=? ?? ��⃗ ??????????????? ? ��⃗= � (?/?) ??? ��⃗ ??????? �? ��⃗?? ? α R1 1ˆ a Rn ? R2 Q1 2ˆ a ? ��⃗?= � [?/?] n a ˆ ?? �? ��⃗? Q2 Qn

9. Example: Find the electric field intensity at the point (2,4,5) m, due to a point charge Q = 2*10-5 C, located at (0,1,2) m. Solution: R Q E πε y x R 3 ˆ 3 ˆ 2 + + = 9 9 4 = + + = R z y x E ) 22 )( 10 85 . 8 ( 4 ∗ π = 3 4 − R 4 ( 0 + = ˆ ) 1 − + − ˆ ) 0 ˆ ) 2 2 ( 5 ( z x y z 22 − ∗ + + 5 ˆ 2 ( ˆ 3 ) ˆ 3 2 10 = = + + x ˆ y ˆ z ˆ ... ... ... − 12 3

10. Electric Field Intensity at a point p (rc, φ, due to line charge ??? �? ��⃗ ?????? ??? ��⃗ ?????? ????? ��⃗ ?????? ? Electric Field Intensity at a Point p (rc, φ, due to Uniform Line Charge Along z-Axis z φ, z) z ?? ��⃗= dE b R dl a y = x ? ? ��⃗= � φ, z) b dz' R ⊕ ⊕ α2 R z' α1 z (rc,φ,z) a z rc y φ x

11. ??? �? ��⃗ ?????? ?? ��⃗= ?? = ???? = ????′ ? ��⃗= ? �????−? �?(?′− ?) ?+ (?′− ?)? ? = �?? ?= ? �????−? �?(?′− ?) �?? ? �? ��⃗=? ��⃗ ?+ (?′− ?)? ?? ��⃗= ????′�? �????−? �?(?′− ?)� ????�?? ?? � ?+ (?′− ?)?� �? �????−? �?�?′−?�� ? ?? ? ��⃗= ??′ ????� ?? � ?� ? �???+ �?′−?� note: ? ?? ? � � = � ?? ?? ? �??+??� ??�??+??�

12. ? ??? −? � � = � ?? ?? ? �??+??� �??+??� ⎡ ? ??′ ?? ⎢ ? �????� ? ��⃗= � ⎢ ?? ???? ?� ? ⎢ �???+ �?′−?� �?′−?�??′ ⎣ ? −? �?� � � ?? ?� ? �???+ �?′−?� (?′− ?) ?? ? ��⃗= ????�? �???? � ?? ???[???+ (?′− ?)?] ? ? + ? �? � � ?? [???+ (?′− ?)?] ? ????�? �?? (? − ?) (? − ?) ?? ? ��⃗= �− �� � [???+ (? − ?)?]?? [???+ (? − ?)?]?? ?? ? +? �? ?? ?? ??� �− ��� [???+ (? − ?)?]?? [???+ (? − ?)?]?? ?

13. ? ????�? �?? [?????+ ?????] +? �? ?? ? ��⃗= ?? [?????− ?????]� ?? ? ?? ? ��⃗= ??????�? �??[?????+ ?????] + ? �? [?????− ?????]� Note: - For infinite line??= ??= ??? So, ? ��⃗= ? �?? Projection point ? �???? Intersection point ?? ??????

14. Electric Field Intensity at a point p (0,0,z) on the axis of a ring charged with uniform ρL of radius a centered at the origin and positioned in (x-y) plane Zˆ E d ( ) z , 0 , 0 R a Yˆ dΦ = φ dl l dl ad = = φ ρ ρ dQ ad l Xˆ ??? �? ��⃗ ?????? ?? ��⃗= ?? = ???? = ???? ′??′ = ?????′ ? ��⃗= −? �???? ′+ ? �?? = −? �??? + ? �?? ′?+ ??= �??+ ?? ? = �??

15. ? �? ��⃗=? ��⃗ ?= −? �??? + ? �?? �??+ ?? ?????′�−? �??? + ? �??� ?? ��⃗= ?? � ????[??+ ??] �−? �???+ ? �??� ?? ? ��⃗= ??? ??′ ????� ?? � ? �??+??� ?? ?? ??? ? ��⃗= −? �?????′ ? �????′ ��� + � � ????[??+ ??]?? ? ? Since, the unit vector ? �??is not a constant unit vector and it is a function of ?′, and since ? �??= ? �?????′+? �?????′ So, ?? ?? ⎡−� ⎤ ? �?????′???′ ? �?????′???′ − � ??? ⎢ ⎥ ? ��⃗= ? ? ⎢ ⎥ ?? ????[??+ ??]?? � ⎢ ⎥ ? �????′ +� ⎣ ⎦ ?

16. ?? ??? ? ��⃗= ? �????′ �� ?? ? ????[??+ ??] ???? ? ��⃗= ? �? � ?? ???[??+ ??] Example: A uniform line charge of infinite extent with ??= ????/? lies on z-axis. Find ? ��⃗ at (6,8,3) m. Solution: ρ = r ˆ l E c πε 2 r 0 c = + = + = 2 2 2 2 6 8 10 rc x y − ∗ 9 r ˆ 2 20 10 ∴ = = r ˆ c ) 36 / E V m c − . 8 ( π 10 2 85 * 10 10

17. Electric field intensity of a surface charge: z d E (0,0,z0) R ρS y rc ds x ??? �? ��⃗ ?????? ?? ��⃗= ?? = ????′ = ???? ′??? ′??′ ? ��⃗= −? �???? ′+ ? �?? ′?+ ?? ? = �??

18. ? �? ��⃗=? ��⃗ ′+ ? �?? ′?+ ?? ?= −? �???? �?? ′??′�−? �???? ′+ ? �??� ?? ′??? ???? ?? ��⃗= � ′?+ ??� ????�?? �??′?�???′ �??′�???′ ? ? ?? ? ? ? ?? ? ?? ????�∫ ? ��⃗= �−? �??�??′+ ∫ ? �????′ �∫ � ∫ ?? ?? � �??′?+??� �??′?+??� Since, the unit vector ? �??is not a constant unit vector and it is a function of ?′, and since ? �??= ? �?????′+? �?????′ So, ′?�??? ?? ? ? ? ?? ′ ?? ???? �?? ? ��⃗= �−? �?????′− ? �?????′�??′ �� � � �??′?+ ??� [?? ? ?? ′]??? ′ ? �????′ + � �� � ?? �??′?+ ??� ? ? Then,

19. ? ?? ′]??? ′ [?? ?? ????�� ? ��⃗= ? �????′ �� � ?? ? ? �??′?+ ??� and since, ? ??′???′ ? ? −? −? ? ∫ � = � �� = � �+ �� ?? ?? [??+??]?? [??]?? �??′?+??� �??′?+??� ? So, ?? ???� −? �+? ? ��⃗= ? �? ?� ?? [??+ ??] ?? ???�? − ? ? ��⃗= ? �? �� ?? �??+??� For infinite surface?→∞ ? ��⃗= ? �? ?? ??? Note: For infinite surface the electric field ? ��⃗= ? �? and in direction normal to the surface and out of it. ?? ???

20. Example: Two infinite uniform sheets of charge ????????? located at ? = ±? as shown in figure. Find the electric field in all regions. Solution: y -1 x=1 x ρ ρ 1 S 2 s Region 1 ρ : ρ s s = = x ˆ x ˆ , E E 1 2 1 2 2 2 ε ε 0 0 [ ρ ]x 1 ε ∴ = + = + ρ ˆ E E E s s 1 2 t 2 1 2 0 Region 2 ρ : ρ s s = − = x ) ˆ x ) ˆ ( ( , E E 1 2 1 2 2 2 ε ε 0 0 [ ]x 1 ε ∴ = − + ρ ρ ˆ E s s t 2 1 2 0

21. Region 3 : ρ ρ s s = − = − x ˆ x ˆ , E E 1 2 1 2 2 2 ε ε 0 0 [ ]x 1 ε ∴ = − + ρ ρ ˆ E s s t 2 1 2 0