- #1
Hawkingo
- 56
- 2
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so in the 2nd page,when the dielectric material is introduced the gauss's law becomes $$\oint _ { S } \vec { E } \cdot \vec { d S } = \frac { ( q - q _ { i } ) } { \epsilon _ { 0 } }$$.But my question is why the ##{ \epsilon _ { 0 } }## is in the equation.Shouldn't it be ##{ \epsilon }(\varepsilon = k \varepsilon _ { 0 })## ?And the formula becomes
$$\oint _ { S } \vec { E } \cdot \vec { d S } = \frac { ( q - q _ { i } ) } { k \varepsilon _ { 0 } }$$
Because
##{ \epsilon _ { 0 } }## is used when the medium is air or vacuum ,but here the medium is dielectric near the gaussian surface,so ##{ \epsilon }## should be used instead of ##{ \epsilon _ { 0 } }## in the gauss's law here.
[PAGE 2]
[PAGE 3]
so in the 2nd page,when the dielectric material is introduced the gauss's law becomes $$\oint _ { S } \vec { E } \cdot \vec { d S } = \frac { ( q - q _ { i } ) } { \epsilon _ { 0 } }$$.But my question is why the ##{ \epsilon _ { 0 } }## is in the equation.Shouldn't it be ##{ \epsilon }(\varepsilon = k \varepsilon _ { 0 })## ?And the formula becomes
$$\oint _ { S } \vec { E } \cdot \vec { d S } = \frac { ( q - q _ { i } ) } { k \varepsilon _ { 0 } }$$
Because
##{ \epsilon _ { 0 } }## is used when the medium is air or vacuum ,but here the medium is dielectric near the gaussian surface,so ##{ \epsilon }## should be used instead of ##{ \epsilon _ { 0 } }## in the gauss's law here.
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