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

Answer from: Quest

radiation

explanation:

heat is transmitted by radiation

Answer from: Quest

we have been given a quadratic function $f(x)=(x-5)^{2} +2$ and we need to restrict the domain such that it becomes a one to one function.

we know that vertex of this quadratic function occurs at (5,2).

further, we know that range of this function is $[2,\infty)$ .

if we restrict the domain of this function to either $(-\infty,5]$ or $[5,\infty)$ , it will become one to one function.

let us know find its inverse.

$y=(x-5)^{2}+2$

upon interchanging x and y, we get:

$x=(y-5)^{2}+2$

let us now solve this function for y.

$(y-5)^{2}=x-2\\ y-5=\pm \sqrt{x-2}\\ y=5\pm \sqrt{x-2}\\$

hence, the inverse function would be $f^{-1}(x)=5+\sqrt{x-2}$ if we restrict the domain of original function to $[5,\infty)$ and the inverse function would be $f^{-1}(x)=5-\sqrt{x-2}$ if we restrict the domain to $(-\infty,5]$ .

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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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