Question

An electric field is given by E vector =(yi^ +xj^ )N/C .the work done in moving a 1C charge from r vector=(2i^ +2j^ )m to r vector = (4i^+j^)m is

Answer 1

first of all, find potential difference between the points.

we know, relation between potential difference and electric field intensity,

$\Delta V=-\int\limits^b_a{E}\,dr$

here, b = (4i + j), a = (2i + 2j)

Let us consider , dr = dx i + dy j

and electric field is given by, $\vec{E}=y\hat{i}+x\hat{j}$

so, potential difference, ∆V = -$\int\limits^b_a{y\hat{i}+x\hat{j}}\,\{dx\hat{i}+dy\hat{j}\}$

= $-\int\limits^b_a{ydx+xdy}$

= $-\int\limits^b_a{d(xy)}$

= $-[xy]^b_a$

now, putting b = (4i + j), a = (2i + 2j)

= -[4 × 1 - 2 × 2 ]

= 0

now workdone = charge × potential difference

= 1C × 0 = 0

hence, workdone in moving a 1C charge from (2i + 2j) to (4i + j) is 0 J