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

This electric field is not possible
Volume charge density is non-uniform.
If a charge particle goes from (3, 4, 5) to (5, 3, 4) then work done by electric force on charge will be positive
Charge enclosed in sphere of unit radius and centered at origin is 4πϵ0.
(A), (B), and (C)
Solution
Solution
Given the electric field
E=xi^+yj^+zk^,we check each statement.
- Statement (A):
The field is conservative since
Its potential is
V=−21(x2+y2+z2)+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.
- Statement (B):
Using Gauss’s law in differential form,
Hence, the volume charge density is
ρ=ϵ0(∇⋅E)=3ϵ0,which is uniform. The statement “volume charge density is non-uniform” is incorrect.
- Statement (C):
The potential corresponding to E is
For points (3,4,5) and (5,3,4),
V(3,4,5)=−21(9+16+25)=−250=−25, V(5,3,4)=−21(25+9+16)=−250=−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.
- Statement (D):
For a sphere of unit radius, the flux is
By Gauss’s law,
ΦE=ϵ0Q⟹Q=4πϵ0.Thus, statement (D) is correct.
Answer: The incorrect statements are (A), (B), and (C).
Explanation (Minimal Core Steps):
- Calculate divergence: ∇⋅E=3 → ρ=3ϵ0 (uniform).
- Potential: V=−21(x2+y2+z2) → same at both points (3,4,5) and (5,3,4) → zero work.
- Flux through a sphere of radius 1: 4π → Q=4πϵ0.