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Question: $\vec{E} = x\hat{i} + y\hat{j} + z\hat{k}$ exists in space then choose INCORRECT statement(s):...

E=xi^+yj^+zk^\vec{E} = x\hat{i} + y\hat{j} + z\hat{k} exists in space then choose INCORRECT statement(s):

A

This electric field is not possible

B

Volume charge density is non-uniform.

C

If a charge particle goes from (3, 4, 5) to (5, 3, 4) then work done by electric force on charge will be positive

D

Charge enclosed in sphere of unit radius and centered at origin is 4πϵ04\pi\epsilon_0.

Answer

(A), (B), and (C)

Explanation

Solution

Solution

Given the electric field

E=xi^+yj^+zk^,\vec{E}=x\,\hat{i}+y\,\hat{j}+z\,\hat{k},

we check each statement.

  1. Statement (A):
    The field is conservative since
×E=0.\nabla\times\vec{E}=\vec{0}.

Its potential is

V=12(x2+y2+z2)+constant.V=-\frac{1}{2}(x^2+y^2+z^2)+\text{constant}.

Even though the potential diverges at infinity, a field defined everywhere (with appropriate charge distribution) is mathematically acceptable. Thus, saying “this electric field is not possible” is incorrect.

  1. Statement (B):
    Using Gauss’s law in differential form,
E=xx+yy+zz=1+1+1=3.\nabla\cdot\vec{E}=\frac{\partial x}{\partial x}+\frac{\partial y}{\partial y}+\frac{\partial z}{\partial z}=1+1+1=3.

Hence, the volume charge density is

ρ=ϵ0(E)=3ϵ0,\rho=\epsilon_0\,(\nabla\cdot\vec{E})=3\epsilon_0,

which is uniform. The statement “volume charge density is non-uniform” is incorrect.

  1. Statement (C):
    The potential corresponding to E\vec{E} is
V=12(x2+y2+z2).V=-\frac{1}{2}(x^2+y^2+z^2).

For points (3,4,5)(3,4,5) and (5,3,4)(5,3,4),

V(3,4,5)=12(9+16+25)=502=25,V(3,4,5)=-\frac{1}{2}(9+16+25)=-\frac{50}{2}=-25, V(5,3,4)=12(25+9+16)=502=25.V(5,3,4)=-\frac{1}{2}(25+9+16)=-\frac{50}{2}=-25.

The potential difference is zero, so the work done by the electric force on a charge is zero, not positive. Thus, statement (C) is incorrect.

  1. Statement (D):
    For a sphere of unit radius, the flux is
ΦE=EdA=(Field magnitude=1, area =4π)=4π.\Phi_E=\oint \vec{E}\cdot d\vec{A}=\text{(Field magnitude=1, area }=4\pi)=4\pi.

By Gauss’s law,

ΦE=Qϵ0    Q=4πϵ0.\Phi_E=\frac{Q}{\epsilon_0}\implies Q=4\pi\epsilon_0.

Thus, statement (D) is correct.

Answer: The incorrect statements are (A), (B), and (C).


Explanation (Minimal Core Steps):

  1. Calculate divergence: E=3\nabla\cdot\vec{E}=3 \rightarrow ρ=3ϵ0\rho=3\epsilon_0 (uniform).
  2. Potential: V=12(x2+y2+z2)V=-\frac{1}{2}(x^2+y^2+z^2) \rightarrow same at both points (3,4,5)(3,4,5) and (5,3,4)(5,3,4) \rightarrow zero work.
  3. Flux through a sphere of radius 1: 4π4\pi \rightarrow Q=4πϵ0Q=4\pi\epsilon_0.