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

A dielectric slab of thickness $$d$$ is inserted in a parallel plate capacitor whose negative plate is at $$x = 3d$$. The slab is equidistant from the plates. The capacitor is given some charge. As $$x$$ goes from $$0$$ to $$3d$$  :
  • the magnitude of the electric field remains the same
  • the direction of the electric field remains the same
  • the electric potential increases continuously
  • the electric potential increases at first, then decreases and again increases
A parallel plate capacitor of capacity $$5 \mu F$$ and plate separation $$6 cm$$ is connected to a $$1V$$ battery and is charged. A dielectric of dielectric constant $$4$$ and thickness $$4 cm$$ is introduced into the capacitor. The additional charge that flows into the capacitor from the battery is:
  • 2$$\mu $$C
  • 3$$\mu $$C
  • 5$$\mu $$C
  • 10$$\mu $$C
Three charges $$Q,+q$$ and $$+q$$ are placed at the vertices of a right angle isosceles triangle as shown. The net electrostatic energy of the configuration is zero, if $$Q$$ is equal to :

125039_a319cbdf683d4b6482b116bbaab249fa.png
  • $$\dfrac { -q }{ 1+\sqrt { 2 } }$$
  • $$\dfrac { -2q }{ 2+\sqrt { 2 } } $$
  • $$-2q$$
  • $$+q$$

Three charges $$Q$$, $$+q$$ and $$+q$$ are placed at the vertices of a right angled isosceles triangle as shown in the figure. The net electrostatic energy of the configuration is zero if $$Q$$ is equal to :


11207_13fd1b26018d44ebb23f9e19e1e61e5a.png
  • $$\dfrac{-q}{1+\sqrt{2}}$$
  • $$\dfrac{-\sqrt{2}q}{1+\sqrt{2}}$$
  • $$-2q$$
  • $$+q$$

An alpha particle of $$5 MeV$$ at a large distance proceeds towards a gold nucleus $$(Z=79)$$ to make a head on collision. The closest distance of approach from the centre of gold nucleus is:
  • $$20 fm$$
  • $$15 fm$$
  • $$10 fm$$
  • $$45 fm$$
Figure shows a hemisphere of charge Q and radius R and a sphere of charge 2Q and radius R. The total potential energy of hemisphere is $$U_{H}$$ and that the sphere is $$U_{S}$$. Then,

72335.jpg
  • $$2U_{H}=U_{S}$$
  • $$2U_{H} < U_{S}$$
  • $$2U_H > U_S$$
  • $$U_{H}=U_{S}$$
An isolated parallel plate capacitor is charged upto a certain potential difference. When a $$3mm$$ thick slab is introduced between the plates then in order to maintain the same potential difference, the distance between the plates is increased by $$2.4mm$$. Find the dielectric constant of the slab. (Assume charge remains constant)
  • 5
  • 10
  • 2.5
  • 7.5
Three identical square metal plates $$M_1, M_2$$ and $$M_3$$ of side 10 cm & 5 mm thick are arranged as shown in figure. The plates are separated by sheets of paper 0.5 mm thick & of dielectric constantThe outer plates are connected together & connected to lower potential while inner plate to higher potential terminal of a battery. Capacitance between the terminals of the battery is : $$(\epsilon _{0} = 8.9 10^{-12} C^{2}/N-m^{2})$$
73207.jpg
  • $$1780pF$$
  • $$890 pF$$
  • $$445 pF$$
  • $$222.5 pF$$
In the figure shown, a parallel plate capacitor has a dielectric of width $$d/2$$ and dielectric constant $$K = 2.$$ The other dimensions of the dielectric are same as that of the plates. The plates $$P_1$$ and $$P_2$$ of the capacitor have area 'A' each. The energy of the capacitor is :

74450.jpg
  • $$\dfrac{\epsilon _{0}AV^{2}}{3d}$$
  • $$\dfrac{2\epsilon _{0}AV^{2}}{d}$$
  • $$\dfrac{3}{2}\dfrac{\epsilon _{0}AV^{2}}{d}$$
  • $$\dfrac{2\epsilon _{0}AV^{2}}{3d}$$
The capacity of a parallel plate condenser without any dielectric is C. If the distance between the plates is doubled and the space between the plates is filled with a substance of dielectric constant $$3$$, the capacity of the condenser becomes:
  • $$\dfrac {3}{4}C$$
  • $$\dfrac {9}{2}C$$
  • $$\dfrac {2}{3}c$$
  • $$\dfrac {3}{2}C$$

112205.jpg
  • Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
  • Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
  • Assertion is correct but Reason is incorrect
  • Assertion is incorrect but Reason is correct
A parallel plate capacitor is connected to a cell. It's positive plate A and the negative plate B have charges $$+Q$$ and $$-Q$$ respectively. A third plate C identical to A and B with charge $$+Q$$ is  now introduced midway between the plates A and B parallel to them What is the charge on the inner surface of A ?

44943.jpg
  • $$Q$$
  • $$\displaystyle \frac{\mathrm{Q}}{2}$$
  • $$\displaystyle \frac{3\mathrm{Q}}{2}$$
  • $$\displaystyle \frac{\mathrm{Q}}{4}$$
A parallel plate capacitor of area $$A$$ and separation $$d$$ is charged to potential difference $$V$$ and removed from the charging source. A dielectric slab of constant $$K = 5$$, thickness $$d$$ and area $$\frac {A}{3}$$ is inserted, as shown in the figure. Let $$\sigma_1$$ be free charge density at the conductor-dielectric surface and $$\sigma_2$$ be the charge density at the conductor-vacuum surface :

75220_ded1486659a245de941bb360afa89abd.png
  • The electric field have the same value inside the dielectric as in the free space between the plates.
  • The ratio $$\dfrac {\sigma_1}{\sigma_2}$$ is equal to $$\dfrac {1}{5}$$
  • The new capacitance is $$\dfrac {7\epsilon_0A}{3d}$$
  • The new potential difference is $$\dfrac {3}{7}V$$
A dielectric slab of thickness d is inserted in a parallel plate capacitor where negative plate is at $$x = 0$$ and positive plate is at $$x = 3d$$. The slab is equidistant from the plates. The capacitor is given some charge. As x goes from $$0$$ to $$3d$$.
  • The magnitude of electric field remains the same.
  • The direction of electric field remain the same.
  • The electric potential increases continuously.
  • The electric potential increase at first then decrease and again increase.
Two identical capacitors $$1$$ and $$2$$ are connected in series to a battery as shown in figure. Capacitor $$2$$ contains a dielectric slab of dielectric constant $$k$$ as shown. $$Q_1$$ and $$Q_2$$ are the charges stored in the capacitors. Now the dielectric slab is removed and the corresponding charges are $$Q'_1$$ and $$Q'_2$$. 
Then
126551.png
  • $$\displaystyle\frac{Q'_1}{Q_1} = \displaystyle\frac{k+1}{k}$$
  • $$\displaystyle\frac{Q'_2}{Q_2} = \displaystyle\frac{k+1}{2}$$
  • $$\displaystyle\frac{Q'_2}{Q_2} = \displaystyle\frac{k+1}{2k}$$
  • $$\displaystyle\frac{Q'_1}{Q_1} = \displaystyle\frac{k}{2}$$
A point charge -q placed at the point A is
112188.jpg
  • In stable equilibrium along x-axis.
  • In unstable equilibrium along y-axis.
  • In stable equilibrium along y-axis.
  • In unstable equilibrium along x-axis.
A parallel plate capacitor of plate area A and plate separation d is charged to p.d. V and then the battery is disconnected. A slab of dielectric constant K is then inserted between the plates of the capacitor so as to fill the space between the plates. If Q, E and W denote respectively the magnitude of charge on each plate, the electric field between the plates after insertion of dielectric and work done on the system in the process of inserting the slab, then-
  • $$Q=\dfrac {\epsilon_0 AV}{d}$$
  • $$Q=\dfrac {\epsilon_0 KAV}{d}$$
  • $$E=\dfrac {V}{Kd}$$
  • $$W=\dfrac {\epsilon_0AV^2}{2d}(1-\dfrac {1}{K})$$
A parallel plate capacitor of plate area $$A$$ and plate separation $$d$$ is charged to potential difference $$V$$ and then the battery is disconnected. A slab of dielectric constant $$k$$ is then inserted between the plates of the capacitor so as to fill the space between the plates. If $$Q,\space E$$ and $$W$$ denote respectively, the magnitude of charge on each plate, the electric field between the plates (after the slab is inserted) and the work done on the system, in question, in the process of inserting the slab, then:
  • $$Q = \displaystyle\frac{\epsilon_0 A V}{d}$$
  • $$Q = \displaystyle\frac{\epsilon_0 k A V}{d}$$
  • $$E = \displaystyle\frac{V}{k d}$$
  • $$W = -\displaystyle\frac{\epsilon_0 A V^2}{2d}\left(1 - \displaystyle\frac{1}{k}\right)$$
A and B are two points in a closed circuit. The potential difference across the condenser of capacity $$5 \mu F$$ is 
1065909_a10029f07a114d168adabc803a866b0d.PNG
  • 6 V
  • 10 V
  • 16 V
  • 4 V
Figure (i) shows a capacitors of capacitance $$C_1$$. A dielectric of dielectric constant K filling half the volume between the plates are inserted as shown in (ii) and (iii) having capacitances $$C_2$$ and $$C_3$$. Then,
146760_f540721ac58b45b996db406cce8f2467.png
  • $$C_1=C_2=C_3$$
  • $$C_2<C_1\   and\  C_3>C_1$$
  • $$C_2>C_1  \space and \space   C_3>C_1$$
  • $$C_2>C_1  \space and \space C_3< C_1$$
A parallel-plate air capacitor of capacitance $$C_0$$ is connected to a cell of emf $$V$$ and then disconnected from it. A dielectric constant $$K$$, which can just fill the air gap of capacitor, is now inserted in it. Which of the following is incorrect?
  • The potential difference between the plates decreases $$K$$ times.
  • The energy stored in the capacitor decreases $$K$$ times.
  • The change in energy is $$\dfrac{1}{2}C_0\varepsilon ^2 (K-1)$$.
  • The change in energy is $$\dfrac{1}{2}C_0\varepsilon ^2 \left ( 1-\dfrac{1}{k} \right )$$.
A parallel-plate air condenser of plate area A and separation d is charged to potential V and then the battery is removed. Now a slab of dielectric constant k is introduced between the plates. If Q, E and W denote respectively the magnitude of charge on each plate, the electric field between the plates (after introduction of dielectric slab) and work done on the system in the process of introducing the slab, then
  • $$\displaystyle W=\frac{\varepsilon _0 A Vh^2}{2d}(1-1/k)$$
  • $$\displaystyle Q=\frac{\varepsilon _0 K A V}{d}$$
  • $$\displaystyle Q=\frac{\varepsilon _0 A V}{d}$$
  • $$\displaystyle E=\frac{V}{kd}$$
A fully charged capacitors has a capacitance $$'C'$$. It is discharged through a small coil of resistance wire embedded in a thermally insulated block of specific heat capacity $$s$$ and mass $$m$$. If the temperature of the block is raised by $$\Delta T$$, the potential difference $$V$$ across the capacitor is :
  • $$\displaystyle \sqrt {\frac{2mC\Delta T}{s} }$$
  • $$\displaystyle {\frac{mC\Delta T}{s} }$$
  • $$\displaystyle {\frac{ms\Delta T}{C} }$$
  • $$\displaystyle \sqrt {\frac{2ms\Delta T}{C} }$$
What is the value of capacitance?
  • $$22.2 \mu F$$
  • $$2.22 \mu F$$
  • $$222 \mu F$$
  • none of these
Two identical capacitors are connected as shown in given figure, having a charge $$q_0$$. A dielectric slab is introduced between the plates of the capacitor (I) so as to fill the gap, keeping the battery remain connected. The charge on each capacitors now will be :

146655.png
  • $$\displaystyle\frac{2q_0}{\left [ 1+(1/k) \right ]}$$
  • $$\displaystyle\frac{q_0}{\left [ 1+(1/k) \right ]}$$
  • $$\displaystyle\frac{2q_0}{(1+k)}$$
  • $$\displaystyle\frac{2q_0}{(1-k)}$$
For the configuration of media of permitivities $${ \varepsilon}_{o },\ \varepsilon $$ and $$\varepsilon_o$$ between parallel plates each of area $$A$$, as show in figure, the equivalent capacitance is :

155158_0527752b8ad44ec6b59d7c56adde1dc9.png
  • $${ \varepsilon }_{ 0 } A/d$$
  • $${ \varepsilon }_{ 0 }\varepsilon A/d$$
  • $$\cfrac { { \varepsilon }_{ 0 }\varepsilon A }{ d(\varepsilon +{ \varepsilon }_{ 0 }) } $$
  • $$\cfrac { { \varepsilon }_{ 0 }\varepsilon A }{ (2\varepsilon +{ \varepsilon }_{ 0 }) } $$
An uncharged parallel plate capacitor having a dielectric of dielectric constant $$K$$ is connected to a similar air cored parallel plate capacitor charged to a potential $${V}_{0}$$. The two share the charge, and the common potential becomes $$V$$. The dielectric constant $$K$$ is :
  • $$\cfrac { { V }_{ 0 } }{ V } -1$$
  • $$\cfrac { { V }_{ 0 } }{ V } +1$$
  • $$\cfrac { V }{ { V }_{ 0 } } -1$$
  • $$\cfrac { V }{ { V }_{ 0 } } +1$$
A fully charged capacitor has a capacitance '$$C$$'. It is discharged through a small coil of resistance wire embedded in a thermally insulated block of specific heat capacity '$$s$$' and mass '$$m$$'. If the temperature of the block is raised $$'\Delta T'$$, the potential difference '$$V$$' across the capacitance is :
  • $$\dfrac {mC\Delta T}{s}$$
  • $$\sqrt {\dfrac {2mC\Delta T}{s}}$$
  • $$\sqrt {\dfrac {2ms\Delta T}{C}}$$
  • $$\dfrac {ms\Delta T}{C}$$
In the following diagram, the charge and potential difference across $$ 8 \mu F $$ capacitance will be respectively (Approximately )
1247486_7cf50d86d5ff45ff881e4dac16c8d8ed.png
  • $$ 320 \mu C, 40 V $$
  • $$ 420 \mu C, 50 V $$
  • $$ 214 \mu C, 27 V $$
  • $$ 360 \mu C, 45 V $$
Two parallel plate capacitors of capacitances $$C$$ and $$2C$$ are connected in parallel and charged to a potential difference $$V$$. The battery is then disconnected, and the region between the plates of $$C$$ is filled completely with a material of dielectric constant $$K$$. The common potential difference across the combination becomes :
  • $$\cfrac {2V}{K+2}$$
  • $$\cfrac {V}{K+1}$$
  • $$\cfrac {3V}{K+3}$$
  • $$\cfrac {3V}{K+2}$$
The plates of a parallel plate capacitor are charged upto $$100V$$. now, after removing the battery, a $$2mm$$ thick plate is inserted between the plates. Then, to maintain the same potential difference, the distance between the capacitor plates is increased by $$1.6mm$$. The dielectric constant of the plate is :
  • $$5$$
  • $$1.25$$
  • $$4$$
  • $$2.5$$
A dielectric slab fills the lower half of a parallel plate capacitor as shown in figure :
(Take plate are as $$A$$)

155243_2c1d9fc893f645b8affb28b037547f1c.png
  • equivalent capacity of the system is $$\big (({ \varepsilon }_{ 0 }A/2d)(1+K)\big )$$
  • the net charge of the lower half of the left hand plate is $$1/K$$ times the charge on the upper half of the plate
  • net charges on the lower and upper halves of the left hand plate are different
  • net charge on the lower hand of the left hand plate is $$\cfrac { K{ \varepsilon }_{ 0 }A }{ 2d } \times V$$
An electric charge $$10^{-3}\mu C$$ is placed at the origin (0, 0) of X-Y co-ordinate system. Two points A and B are situated at $$(\sqrt 2, \sqrt 2)$$ and (2, 0) respectively. The potential difference between the points A and B will be :
  • 4.5 volt
  • 9 volt
  • zero
  • 2 volt
The electric field in the dielectric slab is:

214500_e04f7b289fb94f3eb2979370916dcc08.png
  • $$\dfrac {V}{Kd}$$
  • $$\dfrac {KV}{d}$$
  • $$\dfrac {V}{d}$$
  • $$\dfrac {KV}{t}$$
An air capacitor of capacity $$C=10\mu F$$ is connected to a constant voltage battery of $$12$$ volt. Now the space between the plates is filled with a liquid of dielectric constant $$5$$. The (additional) charge that flows now from battery to the capacitor is :
  • $$120 \mu C$$
  • $$600 \mu C$$
  • $$480 \mu C$$
  • $$24 \mu C$$

214049_7221ec18934a496fa9c0be2a3f6fb6e1.png
  • Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
  • Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
  • Assertion is correct but Reason is incorrect
  • Assertion is incorrect but Reason is correct
Figure shows a parallel plate capacitor with plate area $$A$$ and plate separation $$d$$. A potential difference is being applied between the plates. The battery is then disconnected and a dielectric slab of dielectric constant $$K$$ is placed in between the plates of the capacitor as shown.
Now, answer the following questions based on above information. The electric field in the gaps between the plates and the dielectric slab will is :

214500_e04f7b289fb94f3eb2979370916dcc08.png
  • $$\dfrac {\varepsilon AV}{d}$$
  • $$\dfrac {V}{d}$$
  • $$\dfrac {KV}{d}$$
  • $$\dfrac {V}{d-t}$$
Two parallel plate capacitors of capacitances $$C$$ and $$2C$$ are connected in parallel and charged to a potential difference $$V$$. The battery is then disconnected and the region between the plates of the capacitor $$C$$ is completely filled with a material of dielectric constant $$K$$. The potential difference across the capacitors now becomes :
  • $$\dfrac {3V}{K+2}$$
  • $$KV$$
  • $$\dfrac {V}{K}$$
  • $$\dfrac {3}{KV}$$
A parallel-plate air capacitor of capacitance $$245 pF$$ has a charge of magnitude $$0.148\mu C$$ on each plate. The plates are $$0.328 mm$$ apart. What is the surface charge density on each plate?
  • $$8.7\mu C/m^2$$
  • $$16.3\mu C/m^2$$
  • $$8.7 mC/m^2$$
  • $$16.3 pC/m^2$$
A $$ 40 \mu F $$ capacitor in defibrillator is charged to 3000 V. the energy stored in the capacitor is sent through the patient during a pulse of duration 2 ms. the power delivered to patient is :
  • 45 kW
  • 180 kW
  • 90 kW
  • 360 kW
Two unlike charges of magnitude q are separated by a distance 2d. The potential at a point midway between them is
  • zero
  • $$\dfrac{1}{4 \pi {\epsilon}_{0}}$$
  • $$\dfrac{1}{4 \pi {\epsilon}_{0}}$$ . $$\dfrac{q}{d}$$
  • $$\dfrac{1}{4 \pi {\epsilon}_{0}}$$ . $$\dfrac{2q}{d}$$
A parallel plate capacitor of plate area $$A$$ and plate separation $$d$$ is charged to potential difference $$V$$ and then the battery is disconnected. A slab of dielectric constant $$K$$ is then inserted between the plates of capacitor so as to fill the space between the plates. If $$Q, E$$ and $$W$$ denote respectively, the magnitude of charge on each plate electric field between the plates (after the slab is inserted), and work done on the system, in question, in the process of inserting the slab, then which is wrong?
  • $$Q=\dfrac {\epsilon_0AV}{d}$$
  • $$Q=\dfrac {\epsilon_0KAV}{d}$$
  • $$E=\dfrac {V}{Kd}$$
  • $$W=\dfrac {\epsilon_0AV^2}{2d}\left (1-\dfrac {1}{K}\right )$$
A parallel plate capacitor is connected to a battery. The quantities charge, voltage, electric field and energy associated with this capacitor are given by $$Q_0, V_0, E_0$$, and $$U_0$$ respectively. A dielectric slab is now introduced to fill the space between the plates with the battery still in connection. The corresponding quantities now given by $$Q, V, E$$ and $$U$$ are related to the previous ones are :
  • $$Q > Q_0$$
  • $$V > V_0$$
  • $$E > E_0$$
  • $$U < U_0$$
Identical charges -q each are placed at 8 corners of a cube of each side b. Electrostatic potential energy of a charge +q which is placed at the centre of cube will be :
  • $$\dfrac {-4\sqrt 2q^2}{\pi \epsilon_0b}$$
  • $$\dfrac {-8\sqrt 3q^2}{\pi \epsilon_0b}$$
  • $$\dfrac {-4q^2}{\sqrt 3\pi \epsilon_0b}$$
  • $$\dfrac {-8\sqrt 2q^2}{\pi \epsilon_0b}$$
Which capacitor has the largest potential difference between its plates ?
292165.png
  • A
  • B
  • D
  • A and D are the same and larger than B or C
The following operation can be performed on a capacitor:
X connect the capacitor to a battery of emf E.
Y disconnect the battery.
Z reconnect the battery with the polarity reversed.
W insert a dielectric slab in the capacitor.
  • In XYZ (perform X, then Y, then Z) the stored electric energy remains unchanged and no thermal energy is developed
  • The charge appearing on the capacitor is greater after the action XWY than after the action XYW.
  • The electric energy stored in the capacitor is greater after the action WXY than after the action XYW.
  • The electric field in the capacitor after the action XW is the same as that after WX.
Positive charge Q is uniformly distributed throughout the volume of a dielectric sphere of radius R. A point mass having charge +q and mass m is fired towards the centre of the sphere with velocity v from a point A at distance r(r> R) from the centre of the sphere. Find the minimum velocity v so that it can penetrate R/2 distance of the sphere. Neglect any resistance other than electric interaction. Charge on the small mass remains constant throughout the motion.
  • $$\displaystyle \left[\frac{1}{2 \pi \varepsilon_0} \frac{Qq}{Rm} \left(\frac{r-R}{r} + \frac{3}{4}\right) \right]^{1/2}$$
  • $$\displaystyle \left[\frac{1}{2 \pi \varepsilon_0} \frac{Qq}{Rm} \left(\frac{r-R}{r} + \frac{3}{8}\right) \right]^{1/2}$$
  • $$\displaystyle \left[\frac{1}{2 \pi \varepsilon_0} \frac{Qq}{Rm} \left(\frac{r-R}{r} - \frac{3}{8}\right) \right]^{1/2}$$
  • $$\displaystyle \left[\frac{1}{4 \pi \varepsilon_0} \frac{Qq}{Rm} \left(\frac{r-R}{r} + \frac{3}{4}\right) \right]^{1/2}$$
Which list below places the capacitors in order of increasing capacitance ?
292165.png
  • A, B, C, D
  • B, A, C, D
  • B, A, D, C
  • A, B, D, C
A parallel plate capacitor of area A and separation $$d$$ is charged to potential difference $$V$$ and removed from the charging source A dielectric slab of constant $$K = 5$$, thickness $$d$$ and area $$\displaystyle \frac{A}{3}$$ is inserted as shown in the figure Let $$\displaystyle \sigma _{1}$$ be free charge density at the conductor-dielectric surface and $$\displaystyle \sigma _{2}$$ be the charge density at the conductor-vacuum surface :

333005_606c0a731db548c094c145bfca14e527.png
  • the electric field have the same value inside the dielectric as in the free space between the plates
  • the ratio $$\displaystyle \frac{\sigma _{1}}{\sigma _{2}}$$ is equal to $$\displaystyle \frac{1}{5}$$
  • the new capacitance is $$\displaystyle \frac{7\varepsilon _{0}A}{3d}$$
  • the new potential difference is $$\displaystyle \frac{3}{7}V $$
Choose the incorrect statement:-
  • the potential energy per unit positive charge in an electric field at some point is called the electric potential
  • the work required to be done to move a point charge from one point to another in an electric field depends on the position of the points.
  • the potenial energy of the system will increase if a positive charge is moved against of Coulombian force
  • the value of funcdamental charge is not equivalent to the electronic charge
0:0:1


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