Question

1. A line of charge aligned along the z-axis with a length of L has a charge density of de(z) = pz Where is the displacement along the axis and 2 = 0 at the center of the line. B gives the linear charge density per length, determining the charge density at a given location of the line. 2=0 - - a. Discuss the meaning of the linear charge density equation and what it physically means about the amount of charge at different locations along the rod. b. Give the total charge of the line. • Explain first using symmetry arguments. • Then prove your claim by calculating it. c. Give the direction of the electric field a distance z = d d > L/2) along the positive z axis. d. Give the magnitude of the electric field a distance of z = d d > L/2) along the positive z axis. e. Give the direction of the electric field a distance z = d d > L/2) along the negative z axis. f. Give the magnitude of the electric field a distance of z = d (d > L/2) along the negative z axis. • Use symmetry and explanation to write out your solution equation for part f. • Do not use math to find a solution for part f.

Answer 1

solution: X (2) = pz - 7=+ >Z z=0 Z=! 2=0 Z=+" Jln a linear charge density ý charge per Unit length. A(z)= B2 means that charge per Unit Length depends on z. On the zaris 7 (2) is positive. On - ve z anis A(z) is negative. magnitude of n(z) increases as One moves away from origin as III increases. b. Total charge a=0 Consider two small pieces of same size at 2=+l and a=-1 lag. These piceces have same size. Hence magnitude of charges On there pieces is same. But signs are opposite. Hence sum of these two charges is zero. since the entire rod is sum of such pains

total charge ou rod is zero. element az u located at NIP do= A(z) dz B z dz = +42 22 A+S46 Supz dz 2211 Q (C)-(-:-)) = 0 C. zad (d)) ou +z axis positive charges between z=0 and z=th create electric field along tz aris Negative charges between z=0 and 2 = 14 -

③ create electric field along -ve canis since the charges are nearer to 2=1 They create a field of larger magnitude. Heuce Net electric field is along t2 ans d. do = 212)dz - Bzdz dĒ = do Î Ba - Untold-2) ? +42 - B z dz 40told-2) a E = Sdě = pentru z dz (2-2) 2 2 2 to 2 d-2 = u - dz = du 늘 ß { r rid-u) (-du) J ß z r II-d du 416o u u 29 - 41€o dth и? dth 2 e. z=d (d>44) on ve z-anis since the charges are closer to e=d. net electric field is along -ve z anis.

> N | fe: [-17 ) E (TC Consider Ĕ at Z=+d substitute ž for (auswer for parted) 3 and -d ford. )ام 三. () E-se() | - [( )。 =) =) S} yo Jo