An infinite sheet carrying a uniform surface charge density `sigma` lies on the xy-plane. The work done to carry a charge q from point `vec(A) = a(hat(i)+2hat(j)+3hat(k))` to the point `vec(B) = a(hat(i)-2hat(j)+6hat(k))` (where a is a constant with the dimension of length and `epsilon_(0)` is the permittivity of free space) is

Options

A. `(3 sigma aq)/(2epsilon_(0))`
B. `(2 sigma aq)/(epsilon_(0))`
C. `(5 sigma aq)/(2epsilon_(0))`
D. `(3 sigma aq)/(epsilon_(0))`

Answer

A

Explanation

Distance between the points , `vec(r ) = vec(B) - vec(A) = a(-4hat(j)+3hat(k))`
Field intensity sheet carrying a uniform surface charge density `sigma` lies on the xy-plane,
`vec(E ) = (sigma)/(2epsilon_0)hat(k)`
`:.` Electric force acting on the charge `q , vec(f) = qvec(E ) = (qsigma)/(2 epsilon_0)hat(k)`
`:.` Work done, `W = vec(F) cdot vec(r) = (q sigma)/(2epsilon_0) hat(k) a(-4hat(j)+3hat(k))=(3q sigma a)/(2epsilon_0)`.