If $$Q$$ charge is given to a spherical sheet of radius $$R$$, the energy of the system is

$$\dfrac {Q^{2}}{8\pi \epsilon_{0}R}$$

$$\dfrac {Q^{2}}{4\pi \epsilon_{0}R}$$

$$\dfrac {Q^{2}}{15\pi \epsilon_{0}R}$$

Consider a cube as shown in the figure-I; with uniformly distributed charge in its entire volume. Intensity of electrical field and potential at one of its vertex P are$$ E_0$$ and $$V_o $$ respectively. A portion of half the size (half edge length) of the original cube is cut and removed as shown in the figure-II. Find modulus of electric field and potential at the point P in the new structure.

$$ \dfrac {E_0}{2}$$ and $$\dfrac {3 V_0}{4} $$

$$ \dfrac {3E_0}{4}$$ and $$\dfrac { V_0}{2} $$

$$ \dfrac {3E_0}{4}$$ and $$\dfrac {7 V_0}{8} $$

$$ \dfrac {7E_0}{8}$$ and $$\dfrac {7 V_0}{8} $$

MCQ Questions for Class 12 Medical Physics Electrostatic Potential And Capacitance Quiz 16

A body of mass one gram and carrying a charge

10^{- 8} C

${ 10 }^{ -8 }C$ passes through two points P and Q. The electrostatic potential at Q is 0V. The velocity of the body at Q is

0.2 {m s}^{- 1}

$0.2{ ms }^{ -1 }$ and at P is

\sqrt{0.028} m s^{- 1}

$\sqrt { 0.028 } m{ s }^{ -1 }$ . The potential at P is

0%
150V
0%
300V
0%
600V
0%
900V

Three uncharged capacitors of capacities

C_{1}, C_{2}

$C_1, C_2$ and

C_{3}

$C_3$ are connected as shown in the figure to one another and the points A, B and C are at potentials

V_{1}, V_{2}

$V_1, V_2$ and

V_{3}

$V_3$ respectively. Then the potential at O will be?

0%
$\dfrac{V_1C_1+V_2C_2+V_3C_3}{C_1+C_2+C_3}$
0%
$\dfrac{V_1+V_2+V_3}{C_1+C_2+C_3}$
0%
$\dfrac{V_1(V_2+V_3)}{C_1(C_2+C_3)}$
0%
$\dfrac{V_1V_2V_3}{C_1C_2C_3}$

A proton and an

α - p a r t i c l e

$\alpha -particle$ enter a uniform magnetic filed perpendicular with the same speed. If proton takes 20

μ s

$\mu s$ to make 5 revolutions, then the periodic time for the

α - p a r t i c l e

$\alpha -particle$ would be

0%
5 $\mu s$
0%
8 $\mu s$
0%
10 $\mu s$
0%
16 $\mu s$

Three charges

- q, + q

$-q, +q$ and

Q

$Q$ are placed at the corners of a right angled isosceles triangles as shown in the figure. If the net potential energy of the system is zero then value of

Q

$Q$ is

0%
$-2q$
0%
$\dfrac{-2q}{\sqrt{2} + 2}$
0%
$\dfrac{-q}{\sqrt{2} + 1}$
0%
$-q$

The distance between the plates of a parallel plate capacitor is

1 m m

$1\ mm$ . What must be the area of the plate of the capacitor if the capacitance is to be

1.0 μ F

$1.0\mu F$ ?

0%
$102.5\ m^{2}$
0%
$205.9\ m^{2}$
0%
$112.9\ m^{2}$
0%
$302.9\ m^{2}$

A point charge

q = 1 C

$q = 1\ C$ and mass

1 k g

$1\ kg$ is projected with speed

10 m s^{- 1}

$10\ ms^{-1}$ in the perpendicular direction of uniform electric field

E = 100 V m^{- 1}

$E = 100\ Vm^{-1}$ The value of latus rectum of the path followed by charged particle (in meter) is :

0%
5
0%
3
0%
1
0%
2

The equivalent capacitance between A and B in the circuit shown in figure, is?

0%
C
0%
$2C$
0%
$3C$
0%
$C/2$

Q

$Q$ charge is given to a spherical sheet of radius

R

$R$ , the energy of the system is

0%
$\dfrac {Q^{2}}{8\pi \epsilon_{0}R}$
0%
$\dfrac {Q^{2}}{4\pi \epsilon_{0}R}$
0%
$\dfrac {Q^{2}}{15\pi \epsilon_{0}R}$
0%
None of these

The parallel combination of two air filled parallel plate capacitors of capacitance C and NC is connected to a battery of voltage, V. When the capacitors are fully charged, the battery is removed and after that a dielectric material of dielectric constant K is placed between the two plates of the first capacitor. The new potential difference of the combined system is?

0%
$\dfrac{V}{K+n}$
0%
$V$
0%
$\dfrac{(n+1)V}{(K+n)}$
0%
$\dfrac{nV}{K+n}$

Explanation

After fully charging, battery is disconnected.
Total charge of the system

= C V + n C V

$=CV+nCV$

= (n + 1) C V

$=(n+1)CV$
After the insertion of dielectric of constant K
New potential (common)

V_{C} = \frac{t o t a l c h a r g e}{t o t a l c a p a c i t a n c e}

$V_C=\dfrac{total charge}{total capacitance}$

= \frac{(n + 1) C V}{K C + n C} = \frac{(n + 1) V}{K + n}

$=\dfrac{(n+1)CV}{KC+nC}=\dfrac{(n+1)V}{K+n}$ .
1247377_1614244_ans_d4c8b856d2b0480c827b8a45d26ca36e.png

Two point charges at certain distance apart in air repel each other with a force

F

$F$ . A glass plate is introduced between the charges. The force becomes

F_{1}

${F}_{1}$ , where

0%
${F}_{1}< F$
0%
${F}_{1}= F$
0%
${F}_{1}> F$
0%
data is insufficient

Consider a cube as shown in the figure-I; with uniformly distributed charge in its entire volume. Intensity of electrical field and potential at one of its vertex P are

E_{0}

$E_0$ and

V_{o}

$V_o$ respectively. A portion of half the size (half edge length) of the original cube is cut and removed as shown in the figure-II. Find modulus of electric field and potential at the point P in the new structure.

0%
$\dfrac {E_0}{2}$ and $\dfrac {3 V_0}{4}$
0%
$\dfrac {3E_0}{4}$ and $\dfrac { V_0}{2}$
0%
$\dfrac {3E_0}{4}$ and $\dfrac {7 V_0}{8}$
0%
$\dfrac {7E_0}{8}$ and $\dfrac {7 V_0}{8}$

We wish to obtain a capacitance of

5 μ F

$5\mu F$ , by using some capacitors, each of

2 μ F

$2\mu F$ . Then, the minimum number of capacitors required is?

0%
$3$
0%
$4$
0%
$5$
0%
Not possible

A number of capacitors, each of equal capacitance C, are arranged as shown in Fig. The equivalent capacitance between A and B is?

0%
$n^2C$
0%
$(2n+1)C$
0%
$\dfrac{(n-1)n}{2}C$
0%
$\dfrac{(n+1)n}{2}C$

In the accompanying diagram, if

C_{1} = 3 μ F

$C_1=3\mu F$ ,

C_{2} = 6 μ F

$C_2=6\mu F$ ,

C_{3} = 9 μ F

$C_3=9\mu F$ ,

C_{4} = 12 μ F

$C_4=12\mu F$ ,

C_{5} = 15 μ F

$C_5=15\mu F$ and

C_{6} = 18 μ F

$C_6=18\mu F$ , then the equivalent capacitance between the ends A and B is?

0%
$1.22\mu F$
0%
$5.16\mu F$
0%
$2.25\mu F$
0%
$2.51\mu F$

When a dielectric slab is introduced between the plates of an isolated charged capacitor, it.

0%
Increases the capacitance of the capacitor
0%
Decreases the electric field between the plates
0%
Decreases the amount of energy stored in the capacitor
0%
All of the above

Time rate at which radius r of the balloon changes is best represented by the expression.

0%
$\dfrac{1}{4\pi \varepsilon_0R}$
0%
$\dfrac{1}{4\pi\varepsilon_0R}$
0%
$\dfrac{r}{4\pi\varepsilon_0aR}$
0%
$-\dfrac{r}{4\pi\varepsilon_0aR}$

Explanation

Let initially ballon has charge

Q

$Q$

then

V_{0} = \frac{Q}{4 π \in_{0} a}

$V_0=\dfrac{Q}{4\pi\in_0a}$

when gas starts escaping out, the radius of ballon start decreasing, To maintain the potential of bubble to

V_{0}

$V_0$

electros from earth starts moving towards the bubble.

As a result of that a current i is setup.

Potential of earth is taken as O.

The value of current is proportional to time rate at which radius of balloon decreases.

\frac{d r}{d t} = i = \frac{V_{0} - 0}{R}

$\dfrac{dr}{dt}=i=\dfrac{V_0-0}{R}$

\Rightarrow \frac{d r}{d t} = \frac{Q}{4 π \in_{0} a R}

$\Rightarrow \dfrac{dr}{dt}=\dfrac{Q}{4\pi\in_0a R}$

thus

\Rightarrow \frac{d r}{d t} \propto \frac{1}{4 π \in_{0} R}

$\Rightarrow \dfrac{dr}{dt}\propto \dfrac{1}{4\pi\in_0R}$

Hence option B is correct
2070757_1732291_ans_48a977332b7d4951a394c1b515d092b2.png

Three identical capacitors, each of capacitance C, are connected in series with a battery of e.m.f. V and get fully charged. Now, the battery is removed and the capacitors are connected in parallel with positive terminals at one point and negative terminals at other point. Then, the common potential will be?

0%
V
0%
$3$ V
0%
$V/3$
0%
Zero

The charged capacitors are disconnected from the line and from each other, and are now reconnected with terminals of like sign together. Find the final charge on each capacitor and voltage across each capacitor.

0%
Charge on capacitors: $\left(\dfrac{1400}{3}\right)\mu C$ and $\left(\dfrac{3200}{3}\right)\mu C$ ; potential difference across each capacitor: $\left(\dfrac{1600}{3}\right)V$
0%
Charge on capacitors: $\left(\dfrac{1600}{3}\right)\mu C$ and $\left(\dfrac{3200}{3}\right)\mu C$ ; potential difference across each capacitor: $\left(\dfrac{1600}{3}\right)V$
0%
Charge on each capacitor is $1600\mu C$ and potential difference across each capacitor is $800$ V
0%
Charge and potential difference across each capacitor are zero

In the given network of capacitors as shown in Fig., given that

C_{1} = C_{2} = C_{3} = 400 p F

$C_1=C_2=C_3=400 pF$ and

C_{4} = C_{5} = C_{6} = 200 p F

$C_4=C_5=C_6=200 pF$ . The effective capacitance of the circuit between X and Y is?

0%
$810$ pF
0%
$205$ pF
0%
$600$ pF
0%
$410$ pF

Two-point charges

4 μ C

$4 \mu C$ and

- 2 μ C

$-2 \mu C$ are separated by a distance of 1 m in the air. At what point in between the charges and on the line joining the charges, is the electric potential zero?

0%
In the middle of the two charges
0%
$1/3 m$ from $4 \mu C$
0%
$1/3 m$ from $-2 \mu C$
0%
Nowhere the potential is zero

A dielectric slab fills the lower half region of parallel-plate capacitor as shown in Fig. [Take plate area as A]

0%
Equivalent capacity of the system is $\dfrac{\varepsilon_0A}{2d}(1+K)$
0%
The net charge of lower half of the left hand plate is $1/K$ times the charge on upper half of the plate
0%
Net charges on lower and upper halves of left hand plate are different
0%
Net charge on lower half of left hand plate is $\dfrac{K\varepsilon_0A}{2d}\times V$

A photographic flash unit consists of a xenon-filled tube. It gives a flash of average power

2000

$2000$ W for

0.04

$0.04$ s. The flash is due to discharge of a fully charged capacitor of

40 μ F

$40\mu F$ . The voltage to which it is charged before a flash is given by the unit is?

0%
$1500$ V
0%
$2000$ V
0%
$2500$ V
0%
$3000$ V

The equivalent capacitance across AB is?

0%
$8\mu F$
0%
$12\mu F$
0%
$4\mu F$
0%
$24\mu F$

The plates of a parallel-plate capacitor have an area of

90 c m^{2}

$90 cm^2$ each and are separated by

2

$2$ mm. The capacitor is charged by connecting it to a

400

$400$ V supply. Then the energy density of the energy stored (in

J m^{- 3}

$Jm^{-3}$ ) in the capacitor is(Take

ε_{0} = 8.8 \times 10^{- 12} F m^{- 1}

$\varepsilon_0=8.8\times 10^{-12} Fm^{-1}$ )

0%
$0.113$
0%
$0.117$
0%
$0.152$
0%
None of these

Solve the question if the dielectric materials were filled as shown in Fig.

Take

C = \frac{A}{ϵ_{0} d}

$C= \dfrac{A}{\epsilon_0 d}$

0%
$2.5 C$
0%
$C$
0%
$1.5 C$
0%
None of these

If the charged capacitors are reconnected with terminals of opposite sign together, find the final charge and voltage across each capacitor.

0%
Charge on capacitors: $\left(\dfrac{1400}{3}\right)\mu C$ and $\left(\dfrac{3200}{3}\right)\mu C$ , potential difference across each capacitor: $\left(\dfrac{1600}{3}\right)V$
0%
Charge on capacitors: $\left(\dfrac{1600}{3}\right)\mu C$ and $\left(\dfrac{3200}{3}\right)\mu C$ ; potential difference across each capacitor: $\left(\dfrac{1600}{3}\right)V$
0%
Charge on each capacitor is $1600\mu C$ and potential difference across each capacitor is $800$ V
0%
Charge and potential difference across each capacitor are zero

Explanation

1200 - \frac{q}{1} - \frac{q}{2} = 0

$1200-\dfrac{q}{1}-\dfrac{q}{2}=0$ or

q = 800 μ C

$q=800\mu C$

Charge on each capacitor is 800 μ C

$\text{Charge on each capacitor is } 800\mu C$

V_{1} = \frac{q}{C_{1}} = \frac{800}{1} = 800 V

$V_1=\dfrac{q}{C_1}=\dfrac{800}{1}=800V$

V_{2} = \frac{q}{C_{2}} = \frac{800}{2} = 400 V

$V_2=\dfrac{q}{C_2}=\dfrac{800}{2}=400V$

$q{'_1}+q{'_2}=800-800=0....(i)$

$\text{Also},\dfrac{q{'_1}}{1}-\dfrac{q{'_2}}{2}=0$

$\text{From equation(ii)}, q{'_1}=q{'_2}=0$

$\text{Therefore, potential difference across each capacitor is zero}$

1873217_1751214_ans_f0881538f4c140ebb11129874fb6ab37.jpeg

The electric potential at O is

0%
$\frac { \sqrt {2} q}{ \pi \epsilon_0 a }$
0%
$\frac { \sqrt {3} q}{ \pi \epsilon_0 a}$
0%
$\frac {q}{ \pi \epsilon_0 a}$
0%
$Zero$

CBSE Questions for Class 12 Medical Physics Electrostatic Potential And Capacitance Quiz 16 - MCQExams.com

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Practice Class 12 Medical Physics Quiz Questions and Answers

CBSE Questions for Class 12 Medical Physics Electrostatic Potential And Capacitance Quiz 16 - MCQExams.com

0:0:3

Practice Class 12 Medical Physics Quiz Questions and Answers

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