In a region, the potential is respresented by `V(x, y, z) = 6x - 8xy - 8y + 6yz`, where `V` is in volts and `x, y, z` are in meters. The electric force experienced by a charge of `2` coulomb situated at point `(1, 1, 1)` is

Options

A. `6sqrt(5)N`
B. 30 N
C. 24 N
D. `4sqrt(35) N`

Answer

D

Explanation

Here, V(x,y,z) = 6x - 8xy - 8y + 6yz
The x, y and z components of electric field `vec(E)` are
`E_(x) = -(del V)/(del x) = -(del)/(del x)(6x - 8xy - 8y + 6yz)`
`= -6-8y) = -6 + 8y`
`E_(y) = -(del V)/(del y) = -(del)/(del y)(6x - 8xy - 8y + 6yz)`
`= -(8x - 8 + 6z) = 8x + 8 - 6z`
`E_(z) = -(del V)/(del z) = -(del)/(del z)(6x - 8xy - 8y + 6yz) = -6y`
`:. vec(E) = E_(x)hat(i) + E_(y)hat(j) + E_(z)hat(k) = (-6 + 8 y)hat(i) + (8x + 8 - 6z)hat(j) - 6y hat(k)`
At point (1,1,1)
`vec(E) = (-6 + 8)hat(i) + (8+8-6)hat(j) - 6hat(k) = 2hat(i) + 10 hat(j) - 6hat(k)`
The magnitude of electric field `vec(E)` is
`|vec(E)| = sqrt(E_(x)^(2) + E_(y)^(2) + E_(z)^(2)) = sqrt((2)^(2) + (10)^(2) + (-6)^(2))`
`= sqrt(140) = 2sqrt(35) NC^(-1)`
Electric force experienced by the charge
`F = qE = 2 C xx 2sqrt(35) NC^(-1) = 4sqrt(35)N`