According to Gauss’s Law, the surface integral of the normal component of electric flux density D over a closed surface containing charge Q is:

\(\frac{{{Q^2}}}{{{\varepsilon _0}}}\)

\(\frac{Q}{{{\varepsilon _0}}}\)

According to Gauss’s Law, the surface integral of the normal compone

1.	According to Gauss’s Law, the surface integral of the normal component of electric flux density D over a closed surface containing charge Q is:
A.	\(\frac{Q}{{{\varepsilon _0}}}\)
B.	ε0 Q
C.	Q
D.	\(\frac{{{Q^2}}}{{{\varepsilon _0}}}\)
Answer» D. \(\frac{{{Q^2}}}{{{\varepsilon _0}}}\)

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According to Gauss’s Law, the surface integral of the normal component of electric flux density D over a closed surface containing charge Q is:

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