Answer:
Point charge will be equal to [tex]q=-4.60\times 10^{-8}C[/tex]
Explanation:
We have given distance of A from point charge [tex]r_{A}=2.70m[/tex]
Distance of point B from point charge [tex]r_{B}=3.80m[/tex]
We know that potential difference due to point charge is given as
[tex]V=\frac{Kq}{r}[/tex]
So potential difference at point A [tex]V_{A}=\frac{Kq}{r_a}[/tex]
And potential difference at point B [tex]V_{B}=\frac{Kq}{r_b}[/tex]
We have given [tex]V_{B}-V_{A}=44volt[/tex]
So [tex]\frac{Kq}{r_b}-\frac{Kq}{r_a}=44[/tex]
[tex]Kq(\frac{1}{r_b}-\frac{1}{r_a})=44[/tex]
[tex]9\times 10^9\times q(\frac{1}{3.80}-\frac{1}{2.70})=44[/tex]
[tex]9\times 10^9\times q\times (0.2631-0.370)=44[/tex]
[tex]q=-4.60\times 10^{-8}C[/tex]
So point charge will be equal to [tex]q=-4.60\times 10^{-8}C[/tex]
