1. Two charges q₁ = -7.35 mC and q₂ = -2.90 mC {q₁ = -3.45 mC and q₂ = +4.80 mC} are placed 76.0 cm apart. If q₁ is to the left of q₂,

(A) what is both the magnitude and direction of the electric force on charge q₁? Ans:{0.258 N to right}

(B) what is both the magnitude and direction of the electric force on charge q₂? Ans:{?}

(C) how many electrons are required to make up a net charge equal to that of q₁? Ans:{2.16x10¹³ e}

(D) At what separation of q₁ and q₂ will the force between them be equal to 38.6 N? Ans:{6.21 cm}

2. Two charges (q₁ = -3.40 mC and q₂ = +5.10 mC) {( q₁ = +2.60 mC and q₂ = -1.75 mC)} are placed in such a way that charge q₁ is 384 cm above the origin on the y-axis and charge q₂ is placed 739 cm to the right of the origin along the x-axis.

(A) What is the magnitude and direction of the electric field at the origin due to charge q₁ alone? Draw a sketch of the direction and angle. Ans:{E₁ = 1,590 N/C, directly downward q = -90^o }

(B) What is the magnitude and direction of the electric field at the origin due to charge q₂ alone? Draw a sketch of the direction and angle. Ans:{{E₂ = 288 N/C to the right q = 0^o relative to x-axis}

(C) What is the magnitude and direction of the electric field at the origin due to both charges q₁ and q₂? Draw a sketch of the direction and angle. Ans:{1,610 N/C, q = -79.7^o relative to x-axis}

(D) What would be the magnitude and direction of the force on a third charge q₃ = -12.6 mC placed at the origin due to charges q₁ and q₂ if these two charge are fixed in place and do not move? Draw a sketch of the direction and angle. Ans:{.0203 N, 100^o relative to x-axis}

Bonus:

(E) Where in space will the magnitude of the electric field equal to zero due to the two original charges alone? Ans:{38.0 m beyond q₂ along the line joining the two charges. This is also the point ( x = 41.1 m, y = -17.5 m) }

3-1. A charge of 44.0 mC is placed on a horizontal rod that is 11.5 cm {2.70 m} long in such a way that the charge is divided equally into two + 22.0 mC charges and placed on the ends of the rod. ( I suggest you place the origin at the left end of the rod and not at the rods center.)

(A) Determine the electric field at a point P that is 19.0 cm to the left {right} of the rod? Ans:{5.51x10⁶ N/C to right}

3-2. A charge of 44.0 mC is placed on a horizontal rod that is 11.5 cm {2.70 m} long in such a way that the charge is distributed evenly along the rod.

(A) Determine the linear charge density l spread along the rod? Ans:{16.3 mC/m}

(B) Determine the electric field at a point P that is 19.0 cm to the left {right} of the rod? Show the steps you use to integrate the charge distribution in finding the electric field. Ans:{+7.21x10⁵ N/C to right}

Note that Tipler does the { } example in section 23-1. What is not clear in his example that r in Column’s law always has to be positive. In particular, r = x_o – x since x_o > x in the {} example but this is not true the nonbracketed problem. If you use r = x_o - x you will get the correct numerical answer but this not correct and it will not work in 3-3.

3-3. A charge of 44.0 mC is placed on a rod that is 11.5 cm {2.70 m} long in such a way that the charge density along the rod increases uniformly from left to right. As a result the linear charge density is given by l = a x when the left end of the rod is at the origin; here a is a constant.

(A) What is the value of the constant a? Ans:{a = 12.1 mC/m²}

BONUS:

(B) Determine the electric field at a point P that is 19.0 cm to the left {right} of the rod? Ans:{+1.25x10⁶ N/C to right}

4-1. A non-conducting sphere of radius 1.50 m has a total charge Q distributed through out the interior of the sphere in such a way that the charge density falls off as r(r) = a/ r where a = 25.5 mC/m ² {r(r) = b/ r² where b = 33.0 mC/m}. (See example 23-7 in Tipler}

(A) Calculate the total charge Q contained within the sphere. Ans:{ Q = 622 mC}

(B) Determine the magnitude and direction of the electric field inside the sphere as a function of the distance from the center of the sphere. Ans:{ E(r) = b/( e_or), outwards}

4-2. A charge of + 120 mC is transferred to the outside of a spherical conducting shell with an inner radius 16.0 cm and an outer radius of 22.0 cm. Next a -48.0 mC {+390 mC} point charge is placed at the center of the conducting shell.

(A) What is the magnitude of the charge on the inner and outer surface of the conducting shell? Ans{Q_inner = -390 mC, Q_outer = +510 mC}

(B) What is the magnitude and direction of the electric field interior to the conducting shell at a distance of 11.0 cm from the center of the shell? Ans:{2.90x10⁸ N/C, outwards}

(C) What is the magnitude and direction of the electric field inside the conducting shell at a distance of 17.0 cm from the center of the shell? Ans:{?}

(D) What is the magnitude and direction of the electric field exterior to the conducting shell at a distance of 31.0 cm from the center of the shell? Ans:{4.78x10⁷ N/C, outwards}

5-1. Two point charges q₁ = 13.5 nC & q₂ = -18.0 nC { q₁ = -88.0 nC & q₂ = 360 nC} are located at the fixed positions of x₁ = 22.0 cm and x₂ = 96.0 cm on the x-axis.

(A) What is the electric potential due to both charges at the origin, x = 0? Ans:{V_total(0) = -225 V}

(B) What is the electric potential due to both charges at x = 50.0 cm? Ans:{4.21 kV}

(C) Where on the x-axis is the electric potential due to both charges equal to zero? Ans:{x = 36.5 cm, -1.94 cm}

(D) How much work is required to move a third charge q₃ = -6.60 mC {13.0 mC} from the origin to x = 50.0 cm along any path (that does not go through one of the charges)? Ans:{57.7 mJ}

BONUS:
(E) If a fourth charge q₄ = +16.0 mC {-11.0 mC} with a mass of 3.90 gm is released from rest at the origin, what is the speed and direction that this charge q₄ will attain when it is very far away from the other two charges? Ans:{1.13 m/s}

6-1. A charge of 44.0 mC is placed on a rod that is 11.5 cm {2.70 m} long in such a way that the charge is spread uniformly along the rod. This is thew same setup as homework problem 3-2.

(A) Calculate the potential due to the charge on the rod at a point P that is 19.0 cm to the left {right} of the rod on the x-axis. Ans:{399 kV}

(B) Calculate the potential due to the charge on the rod at a point S that is 13.0 cm above the left end of the rod. Ans:{547 kV}

(C) How much work will it take to move a 840 mC from the point P to the point S along any path? Ans:{124 J}

(D) If the origin is placed at the left end of the rod, determine the equation of the electric potential along the x-axis to the left {right} of the rod as a function of distance from the origin x, the length of the rod L, and the total charge Q placed on the rod.
Ans:{ }

BONUS:

(E) Using your answer to (D) for the electric potential, calculate the electric field to the left {right} of the rod from the fact that the electric field is the negative of the gradient of the potential and verify that this equation gives the same value as 3-2 (B) at x = -.19 cm { x = L + .19 cm}.
Ans:{ , E(2.89 m) = +7.21x10⁶ N/C}