Question

​A ​very ​long ​insulating ​cylinder ​of ​charge ​of ​radius ​2.50 ​cm ​carries ​a ​uniform ​linear ​density ​of ​15.0 ​nC/m. ​(a) Use ​Gauss’s ​law ​to ​find ​an ​expression ​for ​the ​electric ​field ​a ​short ​distance ​r ​from ​the ​cylinder. ​(b) ​If ​you ​put one ​probe ​of ​a ​voltmeter ​at ​the ​surface, ​how ​far ​from ​the ​surface ​must ​the ​other ​probe ​be ​placed ​so ​that ​the voltmeter ​reads ​175

Answer 1

Answer:

r2 = 2.401557  cm

distance = 0.10 cm

Explanation:

given data

​radius ​= 2.50 ​cm

​density ​= ​15.0 ​nC/m

voltmeter ​read =  ​175

solution

we know here potential difference that is express as

ΔV = $\frac{\lambda }{2\pi \epsilon _o} ln\frac{r2}{r1}$     ...........1

so here

$ln\frac{r2}{r1} = 2\pi \epsilon _o \times \frac{\triangle V}{\lambda }$  

as here $\lambda$ is linear charge density  

$\frac{r2}{r1} = e^{2\pi \epsilon _o \times \frac{\triangle V}{\lambda }}$  

r2 = r1 × $e^{2\pi \epsilon _o \times \frac{\triangle V}{\lambda }}$  

r2 = 2.40 × $e^{2\pi 8.85\times 10^{-12} \times \frac{175}{15\times 10^{-6} }}$  

r2 = 2.401557  cm

and

here distance above surface will be

distance = r1 - r2

distance =  2.50 - 2.40

distance = 0.10 cm