Physics, 14.04.2020 21:27 nmoore017

A 10-cm-long thin glass rod uniformly charged to 15.0 nC and a 10-cm-long thin plastic rod uniformly charged to - 15.0 nC are placed side by side, 3.70 cm apart. What are the electric field strengths E1 to E3 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?

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Answer from: ava108

354311 N/C, 275095 N/C, 458528 N/C

Explanation:

E = kQ/[r * √(r² + L/2)²], where

L = 10 cm = 0.1 m

E = kQ/[r * √(r² + 0.05²)]

For the glass rod,

E at 1 cm =

E = k * 15*10^-9 / [0.01 * √(0.01² + 0.05²)]

E = k * 15*10^-9 / (0.01 * 0.051)

E = k * 15*10^-9 / 0.00051

E = (8.99*10^9 * 15*10^-9) / 0.00051

E = 264411 N/C

E at 2 cm

E = k * 15*10^-9 / [0.02 * √(0.02² + 0.05²)]

E = k * 15*10^-9 / (0.02 * 0.054)

E = k * 15*10^-9 / 0.00108

E = [8.99*10^8 * 15*10^-9] / 0.00108

E = 124861 N/C

E at 3 cm

E = k * 15*10^-9 / [0.03 * √(0.03² + 0.05²)]

E = k * 15*10^-9 / (0.03 * 0.0583)

E = k * 15*10^-9 / 0.00175

E = (8.99*10^9 * 15*10^-9) / 0.00175

E = 77057 N/C

For plastic rods

E at (3.7 - 1) cm

E = k * 15*10^-9 / [0.027 * √(0.027² + 0.05²)]

E = k * 15*10^-9 / (0.027 * 0.0568)

E = k * 15*10^-9 / 0.0015

E = (8.99*10^9 * 15*10^-9) / 0.0015

E = 89900 N/C

E at (3.7 - 2) cm

E = k * 15*10^-9 / [0.017 * √(0.017² + 0.05²)]

E = k * 15*10^-9 / (0.017 * 0.0528)

E = k * 15*10^-9 / 0.0008976

E = (8.99*10^9 * 15*10^-9) / 0.0008976

E = 150234 N/C

E at (3.7 - 3) cm

E = k * 15*10^-9 / [0.007 * √(0.007² + 0.05²)]

E = k * 15*10^-9 / (0.007 * 0.0505)

E = k * 15*10^-9 / 0.0003535

E = (8.99*10^9 * 15*10^-9) / 0.0003535

E = 381471 N/C

The net E are

E(net) at p1 = 264411 + 89900

E(net) at p1 = 354311 N/C

E(net) at p2 = 124861 + 150234

E(net) at p2 = 275095 N/C

E(net) at p3 = 77057 + 381471

E(net) at p3 = 458528 N/C

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A 10-cm-long thin glass rod uniformly charged to 15.0 nC and a 10-cm-long thin plastic rod uniformly...

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