Electrostatic Potential and Capacitance Test

Result

Electrostatic Potential and Capacitance Test - 75

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
1 / -0

In an $$LCR$$ circuit as shown below both switches are open initially. Now switch $${S}_{1}$$ is closed, $${S}_{2}$$ kept open. ($$q$$ is charge on the capacitor and $$\tau=RC$$ is capacitive time constant). Which of the following statement is correct?

A
At $$t=\tau,q=CV/2$$

B
At $$t=2\tau,q=CV(1-{e}^{-2})$$

C
At $$t=\dfrac { \tau }{ 2 },q=CV(1-{e}^{-1})$$

D
Work done by the battery is half of the energy dissipated in the resistor.

Solution
Question 2
1 / -0

.

A
$$3\mu F$$

B
$$2\mu F$$

C
$$4\mu F$$

D
$$8\mu F$$

Solution
Question 3
1 / -0

A parallel plate capacitor having plates of area $$S$$ and plate separation $$d$$, has capacitance $${C}_{1}$$ in air. When two dielectrics of different relative permittivities ($${ \epsilon }_{ 1 }=1$$ and $${ \epsilon }_{ 2 }=4$$) are introduced between the two plates as shown in the figure, the capacitance become $${C}_{2}$$. The ratio $$\dfrac { { C }_{ 2 } }{ { C }_{ 1 } }$$ is

A
$$6/5$$

B
$$5/3$$

C
$$7/5$$

D
$$7/3$$

Solution

$$C = 2\varepsilon_0 \dfrac{s}{d};$$ $$C' = 4\varepsilon_0 \dfrac{s}{d}$$

They are in series, their equivalent $$ =\dfrac{4\varepsilon_0 \dfrac{s}{d}\times 2\varepsilon_0 \dfrac{s}{d}}{4\varepsilon_0 \dfrac{s}{d} + 2\varepsilon_0 \dfrac{s}{d} }$$ $$ = \dfrac{4}{3}\varepsilon_0 \dfrac{s}{d}$$

$$C_3 = \dfrac{2\varepsilon_0 s}{2d}$$ $$ = \dfrac{\varepsilon_0 s}{d}$$

$$C_2 = \dfrac{4\varepsilon_0 s}{3d}$$ $$ + \dfrac{\varepsilon_0 s}{d}$$ $$ = \dfrac{7}{3}\varepsilon_0 \dfrac{s}{d}$$

And, $$ C_1 = \varepsilon_0 \dfrac{s}{d}$$

$$ \dfrac{C_2}{C_1}= \dfrac{7}{3}$$
Question 4
1 / -0

The gravitational field strength $$\vec {E}$$ and gravitational potential $$V$$ are related as $$\vec {E} = -\left (\dfrac {\partial V}{\partial x} \hat {i} + \dfrac {\partial V}{\partial y} \hat {j} + \dfrac {\partial V}{\partial z}\hat {k}\right )$$
In the figure, transversal lines represent equipotential surfaces. A particle of mass $$m$$ is released from rest at the origin. The gravitational unit of potential, $$1\overline {V} = 1\ cm^{2}/ s^{2}$$.
X-component of the velocity of the particle at the point $$(4cm, 4cm)$$ is

A
$$4\ cm/s$

B
$$2\ cm/s$$

C
$$2\sqrt {2} \dfrac {cm}{s}$$

D
$$1\ cm/s$$
Question 5
1 / -0

Figure shows four identical metal plate each of area $$A$$. The plates are kept at equal distance $$d$$ successively. A source of emf $$12V$$ is connected between $$x$$ and $$B$$. Potential difference between $$x$$ and $$y$$ is

A
$$10V$$

B
$$8V$$

C
$$6V$$

D
$$4V$$

Solution
Question 6
1 / -0

The diagram shows a small bead of mass $$m$$ carrying charge $$q$$. The bead can freely move on the smooth fixed ring placed on a smooth horizontal plan. In the same plane a charge $$+Q$$ has also been fixed as shown. The potential at the point $$P$$ due to $$+Q$$ is $$V$$. The velocity with which the bead should projected from the point $$P$$ so that it can complete a circle should be greater than

A
$$\sqrt{\dfrac{6qV}{m}}$$

B
$$\sqrt{\dfrac{qV}{m}}$$

C
$$\sqrt{\dfrac{3qV}{m}}$$

D
$$none$$

Solution

Hint:
Potential due to charge $$q$$ at distance $$r$$ from it is given as,
$$V = \dfrac{1}{4 \pi \epsilon_0} \dfrac{q}{r}$$

Potential Energy of charges $$q_1$$ and $$q_2$$ at distance $$r$$ is given as,
$$U = \dfrac{1}{4 \pi \epsilon_0} \dfrac{q_1 q_2}{r}$$

Correct Option: A.

Step 1: Note all the values given and find the condition for which charge can complete the circle.
Mass of small charge on the bead is $$m$$, and charge on it is $$+q$$, it is placed on bead such that it moves on ring smoothly. That is, there is no energy loss by friction.
The charge $$+Q$$ is fixed in the ring as given in the figure.

Charge $$+q$$ and $$+Q$$ both are positive, thus they repel each other. If we somehow bring the charge to the point which is diametrically opposite to point P, the repulsion Force on $$+q$$ will increase and it will itself move back to point P thus completing the circle. So we only need to move the charge to complete semicircle.

Step 2: Find the change in Potential Energy.
Potential due to charge $$+Q$$ at point P is given as,
$$V = \dfrac{1}{4 \pi \epsilon_0} \dfrac{Q}{4a} = qv$$

Potential Energy of charges when $$+q$$ is at point P is givne as,
$$U_i = \dfrac{1}{4 \pi \epsilon_0} \dfrac{Qq}{4a} = qv$$

Potential Energy of charges when $$+q$$ is at point which is diametrically opposite to point P, is
$$U_f = \dfrac{1}{4 \pi \epsilon_0} \dfrac{Qq}{a} $$

$$U_f = 4qV$$

Step 3: Find the velocity of the bead with which it should be projected.
Let us project the bead with velocity $$v$$, and let it stop at the point when it completes semicircle. Since, no external forces are present, by conservation of energy, we have
$$-(U_f - U_i) = KE_f - KE_i$$

$$\Rightarrow -(4qV - qV) = 0 - \dfrac{1}{2}mv^2$$

$$\Rightarrow 3qV = \dfrac{1}{2}mv^2$$

$$\Rightarrow \dfrac{6qV}{m} = v^2$$

$$\Rightarrow v = \sqrt{\dfrac{6qV}{m}}$$

Thus, minimum velocity with which bead should be projected is given as, $$ v = \sqrt{\dfrac{6qV}{m}}$$.
Question 7
1 / -0

The equivalent capacitance across $$A$$ and $$B$$ in the adjoining circuit is:

A
$$\dfrac { C }{ 2 }$$

B
$$C$$

C
$$\dfrac { 3C }{ 2 }$$

D
$$2C$$

Solution
Question 8
1 / -0

The capacity of a parallel plate capacitor formed by the plates of same are A is 0.02$$\mu F$$ with air as dielectric. Now one plate is replaced by a plate of area 2A and dielectric (K = 2) is introduced between the plates, the capacity is :

A
0.04 $$
\mu F
$$

B
0.08$$
\mu F
$$

C
9.91 $$
\mu F
$$

D
2 $$
\mu F
$$

Solution
Question 9
1 / -0

$$4\ \mu F$$ and $$6\ \mu F$$ capacitors are joined in series and $$500\ v$$ are applied between the outer plates of the system. What is the charge on each plate ?

A
$$1\cdot 2\times 10^{3}\ C$$

B
$$6\cdot 0\times 10^{3}\ C$$

C
$$5\cdot 0\times 10^{-3}\ C$$

D
$$2\cdot 0\times 10^{-3}\ C$$

Solution

Since the capacities are connected in series
$$\dfrac{1}{C_{eq}}=\dfrac{1}{C_1}+\dfrac{1}{C_2}$$
$$\dfrac{1}{C_{eq}}=\dfrac{1}{4\mu}+\dfrac{1}{6\mu}$$
$$C_{eq}=\dfrac{12\mu}{5}$$
$$C=\dfrac{Q}{V}$$
$$\dfrac{12\mu}{5}=\dfrac{Q}{500}$$
$$Q=1.2\times 10^{3}$$
Question 10
1 / -0

Which of the following statements are true?

A
equivalent resistance of two resistances in seires is more than each of the individual resistances separately

B
equivalent capacitance of two capacitances in series is less than each the individual capacitances

C
equivalent capacitance of two capacitances in parallel is more than each the individual capacitances separately

D
all of the above

Solution

Let us consider two resistances $$R_{1}, R_{2}$$.
As equivalent resistance in series, $$=R_{1}+R_{2}$$ which is more than $$R_{1}$$ & $$R_{2}$$ respectively.
Capacitance in series$$=\cfrac{1}{C_{1}}+\cfrac{1}{C_{2}}$$ which is less than individual capacitance.
Capacitance in parallel$$=C_{1}+C_{2}$$ which is more than individual capacitance.
Hence, all statements are true.