Question

Two 20cm -long thin glass rodes uniformly chraged to +18nC are placed side by side, 4.0 cm apart. What are the electric field strengths E1, E2 and E3 at distances 1.0cm, 2.0 cm and 3.0 cm to the right of the rod on the left, along the line connecting the midpoints of the two rods?

Part A) Specify the electric field strength E1 ________ value and units

Part B) Specify the electric field strength E2

Part C) Specify the electric field strength E3

Answer 1

The electric field strength at point P distance D from a charged rod along its perpendicular bisector is
E = (2k?/D)sin?,
where k=1/4??0, ?=linear charge density of rod, and ? is half-angle subtended at P by the rod.
(see earlier Answer in Sources.)

The total electric field at point P due to both rods is
E=E1+E2=2k?(sin?1/D1+sin?2/D2).
1 = glass rod, 2= plastic rod. The + sign occurs because the field points away from the +ve charge on the glass rod and towards the -ve charge on the plastic rod.

k=1/4??0=1/(4*3.142*8.85*10^-12)=9.0x10... m/F = 9.0x10^9 Vm/C .

The magnitude of charge density is the same for both rods : ? = 9nC/0.1m= 90nC/m.

2k? = 2x (9.0x10^9)Vm/C x (90x10^-9)C/m = 1620 V.

The separation of the rods is 4.2cm, so the distances of P from each rod are D1=(1.0, 2.0, 3.0)cm and D2=(3.2, 2.2, 1.2)cm.

tan?1= 5/D1 = (5/1.0, 5/2.0, 5/3.0) so ?1 = (1.3734, 1.1903, 1.0304) radians, so
sin?1= (0.9806, 0.9285, 0.8575).

tan?2= 5/D2 = (5/3.2, 5/2.2, 5/1.2) so ?2 = (1.0015, 1.15629, 1.3353) radians, so
sin?2= (0.8423, 0.9153, 0.9724).

sin?1/D1 = (0.98058/0.01, 0.92848/0.02, 0.85749/0.03) m^-1 = (98.06, 46.42, 28.58)m^-1, and
sin?2/D2 = (0.84227/0.032, 0.91532/0.022, 0.97239/0.012)m^-1 = (26.32, 41.61, 81.03)m^-1.

(note change from cm to m for D1 & D2, so that E comes out in V/m; but you can leave them in cm and state field in V/cm if you prefer.)

sin?1/D1 + sin?2/D2 = (124.38, 88.0298, 109.62)m^-1.

E = 1620 V x (124.38, 88.0298, 109.62)m^-1
= (202, 143, 178) kV/m after rounding to 3 sig. figs.