Question

A 10 cm long thin glass rod uniformly charged to +10 nC and a 10 cm long thin plastic rod uniformly charged to -10nC are placed side by side 4 cm apart.

a. Find the electric field 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. (Hint: treat the rods as a uniform line of charges.)

b. If the plastic rod is exchanged for a glass rod of the same size uniformly charged to +10 nC, what is the new electric field at 1.0 cm, 2.0 cm and 3.0 cm?

Answer 1

Given

Length of the rod L = 10 cm = 0.10 m

The magnitude of charge on glass rod Q= 10 x 10^-9 C

Solution

A)

1)

At D = 1 cm = 0.01 m

Electric field due to glass rod

E_g = kQ/D(D+L)

E_g = 9 x 10⁹ x 10 x 10^-9 / 0.01(0.01 + 0.1)

E_g = 8.2 x 10⁴ N/C ( towards right)

Point A is at distance 3 cm from plastic rod

r = 4 -D

r = 4-1 = 3 cm = 0.03 m

Electric field due to plastic rod

E_p = KQ/r(r+L)

E_p = 9 x 10⁹ x 10 x10^-9 / 0.03(0.03 +0.1)

E_p = 2.3 x 10⁴ N/C ( towards right)

Total Field

E = E_g + E_p

E = 8.2 x 10⁴ + 2.3 x 10⁴

E = 10.5 x 10⁴ N/c ( towards right)

2)

At D = 2 cm = 0.02 m

Electric field due to glass rod

E_g = kQ/D(D+L)

E_g = 9 x 10⁹ x 10 x 10^-9 / 0.02(0.01 + 0.2)

E_g = 3.75 x 10⁴ N/C ( towards right)

Point A is at distance 2 cm from plastic rod

r = 4 -D

r = 4-2 = 2 cm = 0.02 m

Electric field due to plastic rod

E_p = KQ/r(r+L)

E_p = 9 x 10⁹ x 10 x10^-9 / 0.02(0.02 +0.1)

E_p = 3.75x 10⁴ N/C ( towards right)

Total Field

E = E_g + E_p

E = 3.75 x 10⁴ + 3.75 x 10⁴

E =7.5 x 10⁴ N/c ( towards right)

3)

At D = 3 cm = 0.03 m

Electric field due to glass rod

E_g = kQ/D(D+L)

E_g = 9 x 10⁹ x 10 x 10^-9 / 0.03(0.01 + 0.3)

E_g = 2.3 x 10⁴ N/C ( towards right)

Point A is at distance 2 cm from plastic rod

r = 4 -D

r = 4-3 = 1 cm = 0.01 m

Electric field due to plastic rod

E_p = KQ/r(r+L)

E_p = 9 x 10⁹ x 10 x10^-9 / 0.01(0.01 +0.1)

E_p = 8.2 x 10⁴ N/C ( towards right)

Total Field

E = E_g + E_p

E = 2.3 x 10⁴ + 8.2 x 10⁴

E = 10.5 x 10⁴ N/c ( towards right)

_____________________________

B)

1)

At D = 1 cm = 0.01 m

Electric field due to glass rod

E_g = kQ/D(D+L)

E_g = 9 x 10⁹ x 10 x 10^-9 / 0.01(0.01 + 0.1)

E_g = 8.2 x 10⁴ N/C ( towards right)

Point A is at distance 3 cm from plastic rod

r = 4 -D

r = 4-1 = 3 cm = 0.03 m

Electric field due to plastic rod

E_p = KQ/r(r+L)

E_p = 9 x 10⁹ x 10 x10^-9 / 0.03(0.03 +0.1)

E_p = 2.3 x 10⁴ N/C ( towards left)

Total Field

E = E_g - E_p

E = 8.2 x 10⁴ - 2.3 x 10⁴

E = 5.9 x 10⁴ N/c ( towards right)

2)

At D = 2 cm = 0.02 m

Electric field due to glass rod

E_g = kQ/D(D+L)

E_g = 9 x 10⁹ x 10 x 10^-9 / 0.02(0.01 + 0.2)

E_g = 3.75 x 10⁴ N/C ( towards right)

Point A is at distance 2 cm from plastic rod

r = 4 -D

r = 4-2 = 2 cm = 0.02 m

Electric field due to plastic rod

E_p = KQ/r(r+L)

E_p = 9 x 10⁹ x 10 x10^-9 / 0.02(0.02 +0.1)

E_p = 3.75x 10⁴ N/C ( towards left)

Total Field

E = E_g - E_p

E = 3.75 x 10⁴ - 3.75 x 10⁴

E =0 N/c

3)

At D = 3 cm = 0.03 m

Electric field due to glass rod

E_g = kQ/D(D+L)

E_g = 9 x 10⁹ x 10 x 10^-9 / 0.03(0.01 + 0.3)

E_g = 2.3 x 10⁴ N/C ( towards right)

Point A is at distance 2 cm from plastic rod

r = 4 -D

r = 4-3 = 1 cm = 0.01 m

Electric field due to plastic rod

E_p = KQ/r(r+L)

E_p = 9 x 10⁹ x 10 x10^-9 / 0.01(0.01 +0.1)

E_p = 8.2 x 10⁴ N/C ( towards left)

Total Field

E = E_p - E_g

E = 8.2 x 10⁴ - 2.3 x 10⁴

E = 5.9 x 10⁴ N/c ( towards left)