CBSE Questions for Class 12 Medical Physics Electrostatic Potential And Capacitance Quiz 11 - MCQExams.com

As we know that equipotential surfaces are perpendicular to the direction electric field
lines. Here electric field is in $$+\hat{z}$$ direction.

So, equipotential surfaces will be the plane perpendicular to $$z$$ axis, i.e.,
along $$x$$ -y, plane, which includes any $$x$$ or $$y$$ axes. So answers $$(b),(c)$$ and $$(d)$$ are verified respectively.

Constant
potential $$\Rightarrow \mathrm{dV}=0$$ so by relation $$E=\dfrac{-d V}{d r},
\mathrm{E}=0$$

i.e., the
E.F. is not uniform discards answer (a) and agree which answer (b).

As potential
may be outside the charge also so there can be no charge inside the region of
constant potential. It verifies answer (c).

If a charge
is placed in outside region, potential difference in region will not be changed
or electric field will not be changed. It makes answer (d) false.

Correct answer: Option C

Hint: A capacitor is defined as an electrical component which stores energy electrostatically in an electric field.

Step 1: Finding final energy

The capacitor originally had the energy U and the work required to pull the insulator out is 4U so the final energy is equal to 5U.

$$W + U = {U_t}$$

$${U_t} = 4U + U$$

$${U_t} = 5U$$

Step 2: Stored energy in terms of capacitance

The capacitance of the insulator with an insulator,

$$C = \dfrac{{A{\varepsilon _0}}}{d}k$$

The formula of the energy stored in the insulator is given by,

$$U = \dfrac{{{q^2}}}{{2C}}$$

Where capacitance is C the area of the insulator is A the dielectric constant is k the distance between the plates is d and k is permittivity of dielectric constant.

When we remove the insulator from the capacitor then the capacitance gets $${C_0}$$

$$U=\dfrac{{{q^2}}}{{2C}}$$

$${C_0} = \dfrac{{A{\varepsilon _0}}}{d}$$

And the stored energy will be,

$${U_t} = \dfrac{{{q^2}}}{{2{C_0}}}$$

Since,

$${U_t} = 5U$$

Step 3: Substituting values

Replacing the value of $$U$$ Ut=5U and $${U_t}$$ Uin the above equation we get,

$$(\dfrac{{{q^2}}}{{2{C_0}}}) = 5(\dfrac{{{q^2}}}{{2C}})$$

$$\dfrac{C}{{{C_0}}} = 5$$

Replacing the value of $$C$$ and $${C_0}$$ in the above relation

$$\dfrac{{\dfrac{{A{\varepsilon _0}}}{d}k}}{{\dfrac{{A{\varepsilon _0}}}{d}}} = 5$$

$$k = 5$$

Capacitance of parallel plate capacitor with air ,$$C_a=\dfrac{\varepsilon _{0}A}{d}    ....(1)$$

Capacitance of parallel plate capacitor with wax ,$$C_w=\dfrac{K\varepsilon _{0}A}{2d}  .....(2)$$

$$(1)/(2),$$

$$ \dfrac{C_a}{C_w}=\dfrac{2}{K}$$

$$\dfrac{1}{2}=\dfrac{2}{K}$$

$$\therefore K=4$$

At time t the level of liquid is,

$$ \dfrac { d }{ 3 }$$$$ -Vt\\ { C }_{ eq }=\dfrac { \dfrac { { \epsilon  }_{ 0 }KA }{ \dfrac { d }{ 3 } -Vt } \cdot \dfrac { { \epsilon  }_{ 0 }A }{ \dfrac { 2d }{ 3 } +Vt }  }{ \dfrac { { \epsilon  }_{ 0 }KA }{ \dfrac { d }{ 3 } -Vt } +\dfrac { { \epsilon  }_{ 0 }A }{ \dfrac { 2d }{ 3 } +Vt }  } =\dfrac { { \epsilon  }_{ 0 }KA }{ K\left( \dfrac { 2d }{ 3 } +Vt \right) +\dfrac { d }{ 3 } -Vt } \\ =\dfrac { 2{ \epsilon  }_{ 0 } }{ \left( \dfrac { 4d }{ 3 } +2Vt \right) +\dfrac { d }{ 3 } -Vt } =\dfrac { 2{ \epsilon  }_{ 0 } }{ \left( \dfrac { 5d }{ 3 } +Vt \right)  } =\dfrac { 6{ \epsilon  }_{ 0 } }{ 5d+3Vt } \\ \tau =R{ C }_{ eq }=\dfrac { 6{ \epsilon  }_{ 0 }R }{ 5d+3Vt } $$