Question

A charge of `-1 C` is at the origin. How much energy would be applied to or released from a `3 C` charge if it is moved from `(-5 ,1 )` to `(2 ,-6 )`?

Answer 1

`W=27*10^9((sqrt 40-sqrt 26)/sqrt(40*26)) J`

Explanation:

`"potential of any point :"V=k*q/d`
`"The potential at the point(-5,1) by the charge of(-1C)"`
`V_A=k*(-1)/sqrt((-5)^2+1'^2)`
`V_A=(-k)/sqrt 26`
`"The potential at the point(2,-6) by the charge of(-1C)"`
`V_B=(k*(-1))/sqrt( 2^2+(-6^2))`
`V_B=(-k)/sqrt( 40)`
`W=q(V_B-V_A)`
`W=3(-k/sqrt40-(-k/sqrt 26))`
`W=3(-k/sqrt 40+k/sqrt 26)`
`W=`
`W=3*9*19^9()`
`W=27*10^9((sqrt 40-sqrt 26)/sqrt(40*26)) J`