In Fig. 22-49, a nonconducting rod of length L=8.15 cm has a charge -q=-4.23 fC uniformly distr

Step 1: Calculate the linear charge density of the rod using the formula $lambda = frac{-q}{L}

In Fig. 22-49, a nonconducting rod of length L=8.15 cm has...

In Fig. $22-49$, a nonconducting rod of length $L=8.15 \mathrm{~cm}$ has a charge $-q=-4.23 \mathrm{fC}$ uniformly distrib- Fig. 22-49 Problem 31. uted along its length. (a) What is the linear charge density of the rod? What are the (b) magnitude and (c) direction (relative to the positive direction of the $x$ axis) of the electric field produced at point $P$, at distance $a=12.0 \mathrm{~cm}$ from the rod? What is the electric field magnitude produced at distance $a=50 \mathrm{~m}$ by $(\mathrm{d})$ the rod and (e) a particle of charge $-q=-4.23 \mathrm{fC}$ that replaces the rod?

Transcript

00:01 For the a part, the linear charge density is as follows, that is lambda is equal to minus q divided by l.

00:08 Now, solve for lambda.

00:13 So, lambda would be equal to minus of minus 4 .23 f3 divided by the value of l, which is given to us as 8 .15 centimeters.

00:27 So this can be written as minus 4 .23 multiplied by 10 raised to power minus 15 columns divided by 8 .15 cm can be written as 8 .15 multiplied by 10 reach to power minus 2 centimeters, which is equals to when we solve this we are left out with minus 5 .19 multiplied by 10 reach to power minus 14 column divided by meters.

00:56 So that is the answer for the first part.

01:02 Now for the b part we have, the magnitude of an electric field is as follows.

01:10 That is e is equals to minus q divided by 4 pi epsilon naught a, l plus a.

01:18 That is the formula.

01:20 Now, for solving for e, we have minus 4 .23 f of c divided by 4 .2 .4 ,000, f of 6, divided by 4.

01:32 Of pi is 3 .14 multiplied by 8 .852 multiplied by 10 lich to power minus 12 which is the value of epsilon not then we have the value for a which is given to us as 12 centimeters multiplied by l which is 8 .15 centimeters plus 12 centimeters now when we solve this further we are left out with minus 4 minus of 4 point over here also we have minus minus of 4 .23 multiplied by 10 raised to power minus 15 column divided by we have over here 4 multiplied by 3 .14 multiplied by 8 .852 multiplied by 10 to power minus 12 multiplied by 12 centimeter can be written as 12 multiplied by 10 reach to power minus 2 centimeters multiplied by 8 .152...

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