An electric field is described as `vec(E) = hat(i)x + hat(k) z`.
The potential difference `V_(AB)` (in volt) between `A(0,0,0)` and `B(2,2,0)` is :

Answer

2

Explanation

From `E = (-dV)/(dr)`
`dV = -vec(E). Vec(dr)`
`int_(B)^(A) dV = - int_(2,2,0)^(0,0,0) (hat(i) x + hat(k)z).(hat(i) dx + hat(j) dy + hat(k) dz)`
`V_(A) - V_(B) = - int_(2,2,0)^(0,0,0) (x dx + z dz)`
`V_(AB) = -[(x^(2))/(2) + (z^(2))/(2)]_(2,2,0)^(0,0,0)`
`= (-1)/(2) [0^(2) - 2^(2)) + (0^(2) - 0^(2))`
= 2 volt