KCET · Physics · Electrostatics
If there are only one type of charge in the universe, then (\( \vec{E} \) Electric field, \( \vec{d} s \rightarrow \) Area vector\()\)
- A \( \oint \vec{E} \cdot \overrightarrow{d_{S}} \neq 0 \) on any surface
- B \( \oint \vec{E} \cdot \vec{d} s \) could not be defined
- C \( \oint \vec{E} \cdot d s=\infty \) if charge is inside
- D \( \oint \vec{E} \cdot d s=0 \) if charge is outside, \( =\frac{q}{\varepsilon_{0}} \) if charge is inside
Answer & Solution
Correct Answer
(D) \( \oint \vec{E} \cdot d s=0 \) if charge is outside, \( =\frac{q}{\varepsilon_{0}} \) if charge is inside
Step-by-step Solution
Detailed explanation
If there are only one type of charge in the universe, then by Gauss Law:
if charge is outside the surface then charge enclosed by gaussian surface is zero. Therefore, \( \oint \vec{E} \cdot d \vec{s}=0 \)
If charge is inside the gaussian surface, then flux \( \phi \) is given by \( \oint \vec{E} \cdot d \vec{s}=\frac{q}{\varepsilon_{0}} \)
if charge is outside the surface then charge enclosed by gaussian surface is zero. Therefore, \( \oint \vec{E} \cdot d \vec{s}=0 \)
If charge is inside the gaussian surface, then flux \( \phi \) is given by \( \oint \vec{E} \cdot d \vec{s}=\frac{q}{\varepsilon_{0}} \)
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