Question

Find whether the given electric fields are possible or impossible electric field:
Ë; = xyî +2 yzj + 3xzk and Ē, = y’î + (2xy + z?)+2 yzk .
(a) Only Ē is a possible electrostatic field.
(b) Both Ē and Ē, are possible electrostatic field.
(c) Only Ē, is a possible electrostatic field.
(d) Both Ē, and Ē, are impossible electrostatic field.​

Answer 1

Given that,

$E=xyi+2yzj+3xzk$

$\bar{E}=y^2i+(2xy+z^2)j+2yzk$

We know that,

The electric fields are possible. If the $\Delta\times E=0$

We need to find the electric field will possible

Using formula for electric field

For equation (I),

$\Delta\times E=(\dfrac{d}{dx}+\dfrac{d}{dy}+\dfrac{d}{dz})\times(xyi+2yzj+3xzk)$

$\Delta\times E=i(\dfrac{d}{dy}(3xz)-\dfrac{d}{dz}(2yz))-j(\dfrac{d}{dx}(3xz)-\dfrac{d}{dz}(xy))+k(\dfrac{d}{dx}(2yz)-\dfrac{d}{dy}(xy))$

$\Delta\times E=i(0-2y)-j(3z-0)+k(0-x)$

$\Delta\times E=-2yi-3zj-kx$

$\Delta\times E\neq 0$

For equation (II),

$\Delta\times \bar{E}=(\dfrac{d}{dx}+\dfrac{d}{dy}+\dfrac{d}{dz})\times(y^2i+(2xy+z^2)j+2yzk)$

$\Delta\times \bar{E}=i(\dfrac{d}{dy}(2yz)-\dfrac{d}{dz}((2xy+z^2))-j(\dfrac{d}{dx}(2y^2)-\dfrac{d}{dz}(y^2))+k(\dfrac{d}{dx}(2xy+z^2)-\dfrac{d}{dy}(y^2))$

$\Delta\times \bar{E}=i(2z-2z)-j(0-0)+k(2y-2y)$

$\Delta\times \bar{E}=0$

Hence,  Only $\bar{E}$ is possible electrostatic field.

(c) is correct option.

Answer 2

Answer:

c is correct answer of this question