Question

Location A is 2.70 m to the right of a point charge q. Location B lies on the same line and is 3.80 m to the right of the charge. The potential difference VB - VA = 44.0 V. What is the magnitude and sign of the charge?

Answer 1

Point charge will be equal to
`q=-4.60* 10^(-8)C`

Step-by-step explanation:

We have given distance of A from point charge
`r_(A)=2.70m`

Distance of point B from point charge
`r_(B)=3.80m`

We know that potential difference due to point charge is given as

`V=(Kq)/(r)`

So potential difference at point A
`V_(A)=(Kq)/(r_a)`

And potential difference at point B
`V_(B)=(Kq)/(r_b)`

We have given
`V_(B)-V_(A)=44volt`

So
`(Kq)/(r_b)-(Kq)/(r_a)=44`

`Kq((1)/(r_b)-(1)/(r_a))=44`

`9* 10^9* q((1)/(3.80)-(1)/(2.70))=44`

`9* 10^9* q* (0.2631-0.370)=44`

`q=-4.60* 10^(-8)C`

So point charge will be equal to
`q=-4.60* 10^(-8)C`