Question

A 10-cm-long thin glass rod uniformly charged to 10.0 nC and a 10-cm-long thin plastic rod uniformly charged to - 10.0 nC are placed side by side, 4.00 cm apart. What are the electric field strengths 1 to Es at distances 1.0 cm, 2.0 cm, and 3.0 cm from the glass rod along the line connecting the midpoints of the two rods? -Part A Specify the electric field strength E1 Express your answer with the appropriate units.

Answer 1

The electric field strength at point P distance D from a charged rod along its perpendicular bisector is

E = (2kλ/D)sinθ,

where k=1/4πε0, λ=linear charge density of rod, and θ 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=2kλ(sinθ1/D1+sinθ2/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=1/4πε0=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 : λ = 10nC/0.1m= 100nC/m.

2kλ = 2x (9.0x10^9)Vm/C x (100x10^-9)C/m = 1800 V.

The separation of the rods is 4cm, so the distances of P from each rod are D1=(1.0, 2.0, 3.0)cm and D2=(3, 2, 1)cm.

tanθ1= 5/D1 = (5/1.0, 5/2.0, 5/3.0) so θ1 = (1.3734, 1.1903, 1.0304) radians, so

sinθ1= (0.9806, 0.9285, 0.8575).

tanθ2= 5/D2 = (5/3, 5/2, 5/1) so θ2 = (1.0304, 1.1903, 1.3734) radians, so

sinθ2= (0.8575, 0.9285, 0.9806).

sinθ1/D1 = (0.98058/0.01, 0.92848/0.02, 0.85749/0.03) m^-1 = (98.06, 46.42, 28.58)m^-1, and

sinθ2/D2 = (0.8575/0.03, 0.9285/0.02, 0.9806/0.01)m^-1 = (28.58, 46.42, 98.06)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.)

sinθ1/D1 + sinθ2/D2 = (126.64, 92.84, 126.64)m^-1.

E = 1800 V x (126.64, 92.84, 126.64)m^-1

= (228, 167, 228) kV/m after rounding to 3 sig. figs.