Question

The electric field strength vector is given by E = 2.5xi+1.5yj. the potential at that point(2,2,1) is consider the potential at origin to be zero

Answer 1

answer : -8 volts

explanation : electric field strength vector is given by, $\vec{E}=2.5x\hat{i}+1.5y\hat{j}$

let us consider that position vector is $\vec{dr}=dx\hat{i}+dy\hat{j}+dz\hat{k}$

we know, $V=-\int\limits^{(2,2,1)}_{(0,0,0)}\vec{E}.\vec{dr}$

$= -\int\limits^{(2,2,1)}_{(0,0,0)}\{2.5x\hat{i}+1.5y\hat{j}\}.\{dx\hat{i}+dy\hat{j}+dz\hat{k}\}$

$= -\int\limits^{(2,2,1)}_{(0,0,0)}\{2.5x.dx+1.5y.dy$

$= -2.5\int\limits^2_0{x}\,dx-1.5\int\limits^2_0{y}\,dy$

= -2.5 [2²/2 - 0²/2] - 1.5 [2²/2 - 0²/2]

= -2.5 [ 2 ] - 1.5 [ 2]

= -5 - 3

= -8 volts