The distance between the plates of a parallel plate condenser is 8,mm and P.D. 120;volts. If a 6,mm thick slab of dielectric constant 6 is

Q: The distance between the plates of a parallel plate condenser is $8\,mm$ and $P.D.$ $120\;volts$. If a $6\,mm$ thick slab of dielectric constant $6$ is introduced between its plates, then

d (d)If nothing is said, it is considered that battery is disconnected. Hence charge remain the same Also ${V_{air}} = \frac{\sigma }{{{\varepsilon _0}}} \times d$ and ${V_{medium}} = \frac{\sigma }{{{\varepsilon _0}}}(d - t + \frac{t}{k})$ $==>$ $\frac{{{V_m}}}{{{V_a}}} = \frac{{(d - t + \frac{t}{k})}}{d}$ $==>$ $\frac{{{V_m}}}{{120}} = \frac{{(8 - 6 + \frac{6}{6})}}{8} $ $==>$ $ V_m = 45\,V$

SECTION - A [PHYSICS - MCQ]

GSEB - English Medium

The distance between the plates of a parallel plate condenser is $8\,mm$ and $P.D.$ $120\;volts$. If a $6\,mm$ thick slab of dielectric constant $6$ is introduced between its plates, then

A
The charge on the condenser will be doubled
B
The charge on the condenser will be reduced to half
CThe $P.D.$ across the condenser will be $320\;volts$
DThe $P.D.$ across the condenser will be $45\;volts$

Medium

STD 11 Science
NEET
STD 12 - 2. Electric potential and capacitance
Physics

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