Question

The electric-field strength vector is given by.
E = (2.5xi +1.5yj). The electric potential at point
(2m, 2m, 1m) is (Considering electric potential at
origin to be zero)
(1) -2V
(2) -4V
(3)-6V
(4) -8V​

Answer 1

answer : option (4)

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

hence, option (4) is correct

Answer 2

Answer:

this will help you.....