The potential field of an electric field E=(yi+xj) is

Question

The potential field of an electric field E=(yi+xj) is (A)  V=-(x+y)+ constant (B)  V= textconstant  (C)  V=-(x2+y2)+cons tan t  (D)  V=-xy+ textconstant

Tardigrade Admin · Accepted Answer

We know that dV=-E.dr =-(yi+xj).(idx+jdy+kdz) =-(ydx+nxdy) dv=-d(xy) Integrating both sides, we get V=-xy+ textconstant

Q. The potential field of an electric field
$E = (y i + x j)$ is

$V = - (x + y) +$ constant

$V = constant$

$V = - (x^{2} + y^{2}) + co n s tan t$

$V = - x y + constant$

We know that $d V = - E . d r$
$= - (y i + x j) . (i d x + j d y + k d z)$
$= - (y d x + n x d y)$
$d v = - d (x y)$
Integrating both sides, we get
$V = - x y + constant$

Q. The potential field of an electric field E=(yi+xj) is

V=−(x+y)+ constant

V=constant

V=−(x2+y2)+constant

V=−xy+constant