CBSE Questions for Class 12 Medical Physics Electrostatic Potential And Capacitance Quiz 10 - MCQExams.com

Two points charges $$4 \mu C$$ and $$-2 \mu C$$ are separated by a distance of 1 m in air. At what point in between the charges and on the line joining the charges, is the electric potential zero? 
  • In the middle of the two charges
  • $$1/3m$$ from $$4\mu C$$
  • $$1/3m$$ from $$-2\mu C$$
  • Nowhere the potential is zero
In a parallel plate capacitor, the region between the plates is filled by a delectric slab. The capacitor is connected to a cell and the slab is taken out.    
  • Some charge is drawn from the cell
  • Some charge is returned to the cell
  • The potential difference across the capacitor is reduced
  • No work is done by an external agent in taking the slab out
The distance between the plates of a parallel plate capacitor is $$d$$. A metal plate of thickness $$d/2$$ is placed between the plates. The capacitance would be then be
  • Unchanged
  • Initial
  • Zero
  • Doubled
A parallel plate capacitor consists of two circular plates each of radius 2 cm, separatrd by a distance of 0.1 mm If Voltage across the plates is at the rate of 5$$\times 10^{13}$$ V/s, then the value of displacement current is
  • 5.50A
  • 5.56$$\times 10^{3}$$A
  • 2.28$$\times 10^{4}$$A
  • 5.56$$\times 10^{2}$$
Two identical particles of mass m carry a charge Q each . Initially one is at rest on a smooth horizontal plane and the other is projected along the plane directly towards first particle a large distance with speed v. The closed distance of approach be 
  • $$\frac{1}{{4\pi {\varepsilon _0}}}\frac{{{Q^2}}}{{mv}}$$
  • $$\frac{1}{{4\pi {\varepsilon _0}}}\frac{{4{Q^2}}}{{m{v^2}}}$$
  • $$\frac{1}{{4\pi {\varepsilon _0}}}\frac{{2{Q^2}}}{{m{v^2}}}$$
  • $$\frac{1}{{4\pi {\varepsilon _0}}}\frac{{3{Q^2}}}{{m{v^2}}}$$
For high frequency a capacitor offer
  • more reactance
  • less reactance
  • zero reactance
  • inifinite reactance
The plates of a parallel plate condenser are pulled apart with a velocity v. If at any instant their mutual distance of separation is x, the magnitude of the time of rate of change of capacity depends on x as follows : 
  • 1/x
  • $$1/x ^{2}$$
  • $$x ^{2}$$
  • x
Equipotential surfaces are shown in fig, then the electric field strength will be 
1224828_1fdfa6cbc47c432ab0320203eec5d923.png
  • 100 Vm-1 along X-axis
  • 100 Vm-1 along Y-axis
  • 200 Vm-1 at an angle $${ 120 }^{ 0 }$$ wirh X-axis
  • 50 Vm-1 at an angle $${ 180 }^{ 0 }$$ wirh X-axis
The resultant capacitance between A and B in the following figure is equal to
1241952_b12ce9035b30437fb7f50021f1785e42.png
  • $$1 \mu F$$
  • $$3 \mu F$$
  • $$2 \mu F$$
  • $$1.5 \mu F$$
P, Q and R are three points in a uniform electric field. The electric potential is
1239393_3498dad3654c48978dae57f8b5049f86.PNG
  • minimum at R
  • minimum at Q
  • minimum at P
  • Same at all three points
A charged capacitor of capacitance C and having charge Q is to be connected with another uncharged capasitor of capasitance C' as shown till the steady state is reached , find the value of C' for heat liberated through the wires to be minimum.
1263382_ffcaa4db0fb44cbca33e3948c28e5305.png
  • zero
  • C
  • C /2
  • 2C
$$75\%$$ of the distance $$d$$ between the parallel plates of a capacitor is filled with a meterial of dielectric constant $$K$$. Find the changed capacitance if original capacitance was $$C_{0}$$.....................
  • $$\left(\dfrac{3K}{K+3}\right) C_{0}$$
  • $$\left(\dfrac{4K}{K+4}\right) C_{0}$$
  • $$\left(\dfrac{4K}{K+3}\right) C_{0}$$
  • $$\left(\dfrac{3K}{K+4}\right) C_{0}$$
Maximum charge on capacitor after switch is closed is
1244168_2e13f898e4b44df7acc285b5b66301e0.png
  • 2 CE
  • 4 CE
  • 6 CE
  • 7 CE
Two capacitors of $$4\ \mu F$$and $$2\ \mu F$$ are connected in series with the battery. If total potential difference across the two capacitors is $$200$$ volts then the ratio  of potential difference across one capacitor to another is
  • $$1:2$$
  • $$2:1$$
  • $$1:4$$
  • $$4:1$$
The capacitor is charged by closing the switch S. The witch is then opened and the capacitor is allowed to discharge. Take $${ R }_{ 1 }={ R }_{ 2 }={ R }_{ 3 }=R$$ (Battery is ideal and connecting wire has negligible resistance). The fraction of the total heat generated, lost in $${ R }_{ 1 }$$ during discharging is :
  • $$\dfrac { 1 }{ 6 } $$
  • $$\dfrac { 1 }{ 3 } $$
  • $$\dfrac { 1 }{ 2 } $$
  • $$\dfrac { 2 }{ 3 } $$
The ratio of charge densities on the surface of two conducting spheres is 3 :lithe radii of t: the spheres are 4 cm and 8 cm the ratio of the electric potential on the surfaces of the sphere 2 is
  • 3 : 4
  • 3 : 1
  • 1 : 3
  • 4 : 9
The energy per unit volume of a dielectric medium is proportional to square of 
  • relative permittivity
  • charge
  • energy
  • electric intensity
The electric potential decreases uniformly from 120 V to 80 V as one moves on the X-axis from $$ x = -1 cm$$ to $$x = +1$$ cm. The electric field at the origin.
  • must be equal to $$20 V/cm$$
  • must be equal to $$2.0 V/cm$$
  • must be greater than $$20 V/cm$$
  • must be less than $$20 V/cm$$
The distance between the plates of a parallel plate capacitor is 3 mm and the potential 1 difference applied is $$3\times10^5 V$$. If an electron travels from one plate to another, the change in its potential energy is
  • $$3\times10^5 eV$$
  • $$900 eV$$
  • $$10^8 eV$$
  • Negligible
If electric intensity $$\overrightarrow { E } $$ is along the X-axis, then the equipotential surfaces are parallel to
  • XOY plane
  • XOZ plane
  • YOZ plane
  • None of these
Let  $$V_0$$ be the potential at the origin in an electric field $$\overset{\rightarrow}{E}=E_x\hat{i}+E_y\hat{j}$$. The potential at the point $$(x,y)$$ is:
  • $$V_0-{_{x}{E}_x}-{_{y}{E}_y}$$
  • $$V_0+{_{x}{E}_x}+{_{y}{E}_y}$$
  • $${_{x}{E}_x}+{_{y}{E}_y}-V_0$$
  • $$\left(\sqrt{x^2+y^2}\right)\sqrt{{E}_{x}^{2}+{E}_{y}^{2}}-V_0$$
The electric potential at a distance of $$3 \,m$$ on the axis of a short dipole of dipole moment $$4\times 10^{-12}$$ coulomb-metre is 
  • $$1.33 \times 10^{-3} V$$
  • $$4 \,mV$$
  • $$12 \,mV$$
  • $$27 \,mV$$
The electric potential due to point charge at a point
  • May be approximately zero
  • May be positive
  • May be negative
  • Any of these
In order to increase the capacity of parallel plate condenser one should introduce between the plates, a sheet of
  • mica
  • tin
  • copper
  • stainless steel
A parallel plate capacitor of area $$60 cm^2$$ and separation 3 mm is charged initially to $$90 \mu C$$. If medium between the plates gets slightly conducting and the p!ate loses the charge initially at rate of $$2.5\times10^{-8}  C/s$$, then what is the magnetic field between the plates?    
  • $$2.5\times10^{-8} T$$
  • $$2.0\times10^{-7} T$$
  • $$1.63\times10^{-11} T$$
  • ZERO
If electric field in  a region is zero, then electric potential in the region
  • Must be zero
  • Must not be zero
  • May be zero
  • None of these
Three capacitors 4, 6 and 12 $$\mu F$$  are connected in series to a 10 V source. The charge on the middle capacitor is
  • 10 $$\mu C$$
  • 20 $$\mu C$$
  • 60 $$\mu C$$
  • 5 $$\mu C$$
In figure two points $$A$$ and $$B$$ are located in a region of electric field. The potential difference $$V_B-V_A$$ is  
1293302_642afed671aa4f8cadaf6d16eaf32c86.png
  • Positive
  • Negative
  • Zero
  • None of these
Two charges $$-5\mu C$$ and $$+10\mu C$$ are placed 20 cm apart. The net electric field at the mid-point between the two charge is 
  • $$4.5\times { 10 }^{ 5 }N/C$$ directed towards $$+10\mu C$$
  • $$13.5\times { 10 }^{ 5 }N/C$$ directed towards $$-5\mu C$$
  • $$13.5\times { 10 }^{ 5 }N/C$$ directed towards $$+10\mu C$$
  • $$4.5\times { 10 }^{ 5 }N/C$$ directed towards $$-5\mu C$$
The change on capacitor $${ c }_{ 1 }$$
1303865_a4fbefe049b94e25b2adb1f97c57efdf.png
  • $$0_{ \mu }c$$
  • $$5_{ \mu }c$$
  • $$10_{ \mu }c$$
  • none of these
In the following circuit the resultant capacitance between A and B is $$1\mu F$$. Find the value of C :


1327539_108de9eab9da458d940873fd36570190.png
  • $$\dfrac { 23 }{ 32 } \mu F$$
  • $$\dfrac { 32 }{ 23 } \mu F$$
  • $$\dfrac { 13 }{ 23 } \mu F$$
  • $$\dfrac { 23 }{ 13 } $$
In the electric network shown, when no current flows through the $$4\, \Omega$$ resistor in the arm $$EB$$, the potential difference between the points $$A$$ and $$D$$ will be : 

1302685_670c3d90227c447d9320b0dd2095635c.PNG
  • $$3$$ V
  • $$4$$ V
  • $$5$$ V
  • $$6$$ V
The radius of the gold nucleus is $$6.6 \times {10^{ - 15}}m$$  and the atomic number is $$79$$. The electric potential at the surface of the gold nucleus is :
  • $$1.7 \times {10^7}V$$
  • $$7.1 \times {10^7}V$$
  • $$1.7 \times {10^9}V$$
  • $$7.1 \times {10^9}V$$
Some charge Q is to be distributed on 3 concentric shell such that surface charge density on each shell is same as shown. Find the electric potential at r(<a) from point O.
1347673_cded0cd5840646d29d116c1616e5e854.png
  • $$\dfrac { Q(a+b-c) }{ 4\pi { \varepsilon }_{ 0 }({ a }^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 }) } $$
  • $$\dfrac { Q(a+b+c) }{ 4\pi { \varepsilon }_{ 0 }({ a }^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 }) } $$
  • $$\dfrac { Q(a+b+c) }{ 4\pi { \varepsilon }_{ 0 }({ a }^{ 2 }-{ b }^{ 2 }+{ c }^{ 2 }) } $$
  • $$\dfrac { Q(a+b-c) }{ 4\pi { \varepsilon }_{ 0 }({ a }^{ 2 }+{ b }^{ 2 }-{ c }^{ 2 }) } $$
The charges $$Q + q$$ and $$+q$$ are placed at the vertices of a right-angle isosceles triangle as shown below. The net electrostatic energy of the configuration is zero, it the value of Q is: 
1331792_b9eb64ca72a54afd84f61aa007cbd8c5.PNG
  • $$\dfrac{-\sqrt{2}q}{\sqrt{2}+1}$$
  • $$-2q$$
  • $$\dfrac{-q}{1+\sqrt{2}}$$
  • $$+q$$
Two condenser of $$2 \mu F$$ and $$4 \mu F$$ are connected in series. The $$p.d$$ of $$1200$$ volt. The $$p.d$$ across $$2 \mu F$$ is -
  • $$400\ V$$
  • $$600\ V$$
  • $$800\ V$$
  • $$900\ V$$
Potential difference between two points is equal to 
  • electric charge /time
  • work done/time
  • work done/charge
  • work done $$\times$$ charge
Two capacitors 'A' of $$3 \mu F  $$ and 'B' of $$ 2\mu F$$ are connected in series. 'A' can withstand a potential difference of 3 KV while 'B' can withstand 5 KV. What maximum potential difference can be applied across the combination?
  • $$8 KV$$
  • $$6 KV$$
  • $$7.5 KV$$
  • $$8.3 KV$$
The charge on a capacitor plate in a circuit, as a function of time, is shown in the figure:
What is the value of current at $$t = 4s$$?
1332120_0c4669cddf464e62bcc1f83f816831bb.PNG
  • $$3\ \mu A$$
  • $$2\ \mu A$$
  • $$Zero$$
  • $$1.5\ \mu A$$
The energy stored in a parallel plate capacitor can be treated as the energy filled. The energy per unit volume, due to the electric field is :
  • $$E^2$$
  • $$\dfrac{1}{2}{\epsilon_0}{E^2}$$
  • $$\dfrac{1}{{2{\epsilon_0}}}{E^2}$$
  • $$\frac{1}{2}{E^2}$$
The potential energy of a 1$$\mathrm { kg }$$ particle free to move alongthe $$x$$ -axis is given by$$V ( x ) = \left( \frac { x ^ { 4 } } { 4 } - \frac { x ^ { 2 } } { 2 } \right) J$$

The total mechanical energy of the particle is $$21 .$$ Then, the maximum speed (in ms $$^ { 1 }$$ ) is
  • $$\dfrac { 3 } { \sqrt { 2 } }$$
  • $$\sqrt { 2 }$$
  • $$\dfrac { 1 } { \sqrt { 2 } }$$
  • 2
 An infinite number of charges 'q' each are placed along the x - axis at x =1, x=4 ,x=8 and so on. If the distance are in meters calculate the electric potential at x=0
  • $$\frac{{3q}}{{8\pi {E_0}}}$$
  • $$\frac { q } { 2 \pi \epsilon _ { 0 } }$$
  • $$\frac { 2 q } { \pi \epsilon _ { 0 } }$$
  • $$\frac { 4 q } { \pi \epsilon _ { 0 } }$$
In the situation shown in figure, what should be the relation between Q and q so that electric potential at centre of the square is zero:
1330482_0485bbc720b24e49a55fae919d02f4a6.PNG
  • $$Q=q$$
  • $$Q=3q$$
  • $$Q=2q$$
  • $$Q=-3q$$
The figure shows an experiment plot for discharging of a capacitor in an $$R-C$$ circuit. The time constant $$t$$ of this circuit lies between 
1377845_fa536e3748c44f979e859708c64e3bc5.png
  • $$0\ and\ 50\ sec$$
  • $$50\ and\ 100\ sec$$
  • $$100\ and\ 150\ sec$$
  • $$150\ and\ 200\ sec$$
Two spheres of capacitance $$3\mu F$$ and $$5\ mu F$$ are charged to $$300\ V$$ and $$500\ V$$ respectibely and are connected together. The common potential in steady state will be :
  • $$400\ V$$
  • $$425\ V$$
  • $$350\ V$$
  • $$375\ V$$
In the capacitor shown in the circuit is charged to 5V and left in the circuit,in 12s the charge on the capacitor will become.
1396513_1ebe96fe877543b9a9a89171a51a8766.png
  • $$ \frac {10}{e} C $$
  • $$ \frac {e}{10} C $$
  • $$ \frac {10}{e^2} C $$
  • $$ \frac {e^2}{10} C $$
The electric potential at a point $$(x,0,0)$$ is given by$$V=\left[\dfrac{1000}{\chi }+\dfrac{1500}{\chi }+\dfrac{500}{\chi }^3\right]$$
then the electric field at $$x =1\, m$$ is (in volt/m)
  • $$-5500 \, \hat{i}$$
  • $$5500 \, \hat{i}$$
  • $$\sqrt{5500 \, \hat{i}}$$
  • Zero
Potential difference between the points B and C of the circuit is 
1393663_508c6295259146acb9e9a0b8225e55d0.PNG
  • $$ \dfrac { ( C_2-C_1 ) }{V} $$
  • $$ \dfrac { ( C_4-C_3 ) }{V} $$
  • $$ \dfrac { ( C_2C_3 - C_1C_4 ) }{ (C_1+C_2+C_3+C_4) } V $$
  • $$ \dfrac {C_1C_4 - C_2C_3 }{ ( C_1+C_2) \times ( C_3+C_4) } V $$
An arc of radius r carries change. The linear density of charge is $$\lambda $$ and the arc subtends a angle $$\frac{\pi }{3}$$ at the center. What is electric potential at the center
  • $$\dfrac{\lambda }{4\varepsilon _{0}}$$
  • $$\dfrac{\lambda }{8\varepsilon _{0}}$$
  • $$\dfrac{\lambda }{12\varepsilon _{0}}$$
  • $$\dfrac{\lambda }{16\varepsilon _{0}}$$
The potential at a point, due to a positive charge of  $$10\mu { C }$$  at a distance of  $$9 m$$  is
  • $$10^{ { 5 } }V$$
  • $$10^{ { 3 } }V$$
  • $$10^{ { 6 } }V$$
  • $$10^{ { 4 } }V$$
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