ELECTRIC field in vacuum \[{{E}_{0}}=\frac{\SIGMA }{2{{\varepsilon }_{0}}}\] When the PLATES are dipped in KEROSENE oil tank, then the electric field between the plate will be \[E=\frac{\sigma }{2{{\varepsilon }_{0}}k}\] \[\because \] \[k>1\] \[\therefore \] \[E<{{E}_{0}}\] So, the electric field between the plates will decrease.

"> ELECTRIC field in vacuum \[{{E}_{0}}=\frac{\SIGMA }{2{{\varepsilon }_{0}}}\] When the PLATES are dipped in KEROSENE oil tank, then the electric field between the plate will be \[E=\frac{\sigma }{2{{\varepsilon }_{0}}k}\] \[\because \] \[k>1\] \[\therefore \] \[E<{{E}_{0}}\] So, the electric field between the plates will decrease.

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Two parallel metal plates having charges \[+Q\] and \[-Q\] face each other at a certain distance between them. If the plates are now dipped in kerosene oil tank, the electric field between the plates will [AIPMT (M) 2010]

NEET Physics in NEET . 8 months ago

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[c] The ELECTRIC field in vacuum \[{{E}_{0}}=\frac{\SIGMA }{2{{\varepsilon }_{0}}}\] When the PLATES are dipped in KEROSENE oil tank, then the electric field between the plate will be \[E=\frac{\sigma }{2{{\varepsilon }_{0}}k}\] \[\because \] \[k>1\] \[\therefore \] \[E<{{E}_{0}}\] So, the electric field between the plates will decrease.

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