A uniform electric field of `100 V//m` is directed at `30^@` with the positive x-axis as shown in figure. Find the potential difference `V_(BA)` if `OA=2`m and `OB=4 m.`

Options

A. `-200(3+sqrt2)`
B. `50(2+sqrt3)`
C. `-100(3+sqrt2)`
D. `-100(2+sqrt3)`

Answer

D

Explanation

This problem can be solved y both the methods discussed avoe.
Mehod 1. Electric field in vector form can be written as `E=(100 cos 30^@hati+100sin30^@hatj)V//m`
`=(50sqrt3hati+50hatj)V//m`
`A-= -(2m,0,0)`
and `B-= -(0,4m,0)`
`:. V_(BA)=V_B-V_A=-int_A^B E.dr`
`=-int_((-2m,0,0)) ^((0,4m,0))(50 sqrt3hati+50hatj).(dxhati+dyhatj+dzhatk)`
`=-[50sqrt3x+50y0_((-2m,0,0))^((0,4m,0))`
`100(2+sqrt3)V`
Method 2. We can also use, `V=Ed`
With the view that `V_AgtV_B` or `V_B -V_A` will be negative
Here, `d_(AB)=OAcos30^@=4xx1/2=(sqrt3+2)`
`:. V_B-V_A=-Ed_(AB)=-100(2+sqrt3)`