Answer:
1.2kV
Explanation:
We are given that
Linear charge density,[tex]\lambda=358nC/m=358\times 10^{-9} C/m[/tex]
[tex]1 nC=10^{-9} C[/tex]
[tex]r_1=5 m[/tex]
[tex]r_2=6 m[/tex]
We have to find the potential difference between poinst 5.0 m and 6.0 m from the wire.
We know that potential difference of wire
[tex]V=2k\lambda ln\frac{r_2}{r_1}[/tex]
Where
[tex]k=9\times 10^9[/tex]
Substitute the values
[tex]V=2\times 9\times 10^9\times 358\times 10^{-9}ln\frac{6}{5}[/tex]
[tex]V=1174.8 V[/tex]
[tex]V=1.17 kV\approx 1.2 kV[/tex]
1kV=1000 V
The potential difference is 1.2kV.
Since it is mentioned that uniform linear charge density of 358 nC/m between the points 5.0 m and 6.0 m from the wire
So,
The potential difference should be
[tex]= 2 \times 9 \times 10^{9} \times 358 \times 10^{-9}\ ln 6\div 5\\\\[/tex]
= 1174.8V
= 1.17kV
= 1.2kV
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