Two 18 cm -long thin glass rods uniformly charged to +10nC are placed side by si

Question

Two 18 cm -long thin glass rods uniformly charged to +10nC are placed side by side, 4.0 cm apart. What are the electric field strengths E1, E2, and E3 at distances 1.0 cm, 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? art A Specify the electric field strength E1. Express your answer to two significant figures and include the appropriate units. Part B Specify the electric field strength E2. Express your answer to two significant figures and include the appropriate units. Part C Specify the electric field strength E3. Express your answer to two significant figures and include the appropriate units.

Explanation / Answer

Electric field Due to each rod is given by E = KQ/x(sqrt x^2 + l/2)^2)

here Q = 10 nC

length l = 0.18 m

k is 9e9 Nm^2/C^2

for part A :

x = 1cm or 0.01 m

E1 = (9*10^9 * 10*10^-9)/(0.01 * sqrt(0.01^2 + (0.18/2)^2))

E1 = 9.93 *10^4 N/C

E1 = 10 *10^4 N/C

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part B:

at x = 0.2 m

this is mid poit between the rods

and as both the charges are of same magnitudes

EF due to charge is same but point in opp Dirtections (

thus ENet at x = 2cm is 0 .

--------------------

part C:

x = 0.3 cm

E3   = (9*10^9 * 10*10^-9)/(0.03 * sqrt(0.03^2 + (0.18/2)^2))

E1 = 3.2 *10^4 N/C

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