Electrostatic Potential and Capacitance Test

Result

Electrostatic Potential and Capacitance Test - 53

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 the given network capacitance $${C_2} = 10\mu F,\,\,{C_1} = 5\mu F$$ and $${C_3} = 4\mu F.$$ The resultant capacitance between $$P$$ and $$Q$$ will be:

A
$$4.7\mu F$$

B
$$1.2\mu F$$

C
$$3.2\mu F$$

D
$$2.2\mu F$$

Solution

FOR IMAGE 01: The Parallel combination are:
$$\left(10+5=15\mu F\right)$$
FOR IMAGE 02: The series combination are:
$$C_s=\cfrac{15\times 4}{15+4}=\cfrac{60}{19}=3.2\mu F$$
Question 2
1 / -0

Calculate the effective capacitance across XY.

A
$$\dfrac{5}{8}C$$

B
$$\dfrac{3}{5}C$$

C
$$\dfrac{5}{3}C$$

D
$$C$$

Solution

FOR IMAGE 01: The parallel combinations are:
$$C_p=C+C=2C$$
FOR IMAGE 02: The series combinations are:
$$C_s=\left[\cfrac{C\times 2C}{C+2C}=\cfrac{2}{3}C\right]$$
FOR IMAGE 03: The parallel combination are:
$$C_{eq}=\cfrac{2}{3}C+C=\cfrac{5}{3}C$$
Question 3
1 / -0

In a series combination of two capacitances $$C'$$ and $$C(C> C)$$ (as shown in the circuit)

A
C' stores more energy than $$C$$

B
$$C$$ stores more energy than $$C'$$

C
potential difference across $$C$$ is more than that across $$C'$$

D
potential difference across $$C'$$ is less than that across $$C$$

Solution

In series combination of capacitor the potential across each capacitor is same and
$$E= \cfrac{1}{2} CV^{2}$$
As $$C >C'$$, so $$C$$ stores more energy.
Question 4
1 / -0

A parallel plate capacitor with air as dielectric is charged to a potential 'V' using a battery. Removing the battery, the charged capacitor is then connected across an identical uncharged parallel plate capacitor filled with wax of dielectric constant 'K' the common potential of both the capacitor is

A
V volts

B
kV volts

C
(k + 1) V volts

D
$$\dfrac{V}{{k + 1}}{\text{volts}}$$

Solution

Capacitance with air as dielectric $$C= \varepsilon_0 \dfrac{A}{d}$$
For identical Capacitor Capacitance with wax as dielectric $$C'= \varepsilon \dfrac{A}{d}$$
$$\therefore C'=KC , \ where \ K=\dfrac{\varepsilon}{\varepsilon_0}$$
When $$C \ and \ C'$$ are connected in parallel charge is distributed and the potential difference is equal for the two capacitors
Resultant capacitance $$C_R=C+C'$$ and the total charge $$Q=CV$$
$$\therefore$$ Common \Potential \V=\dfrac{Q}{C_R}=\dfrac{CV}{C+C'}$$
$$\therefore V=\dfrac{CV}{C+KC}=\dfrac{V}{K+1}$$
Question 5
1 / -0

Find the equivalent capacitance across A and B for the arrangement shown in the figure. All capacitors are of capacitance C.

A
3C/14

B
C/8

C
3C/16

D
none of these
Question 6
1 / -0

The equivalent capacitance between $$A$$ and $$B$$ in the circuit shown is:

A
$$6\mu F$$

B
$$1.5\mu F$$

C
zero

D
$$2\mu F$$

Solution

Here,
There as direct connection between $$A$$ and $$B$$ by resistance.

And resistance makes charge flow through it(i.e current).

Hence, the resistance will act as a bypass to capacitors leading to

zero capacitance in the circuit.
Question 7
1 / -0

The electric field due to the electric potential V = $$(2x^{2}-4x)$$ is

A
$$(4x+4)\hat{i}$$

B
$$(4x-4)\hat{i}$$

C
$$(-4x+4)\hat{i}$$

D
$$(-4x-4)\hat{i}$$

Solution

$$\vec { E } =-\dfrac{dV}{dx}$$

Given, $$V={ 2x }^{ 2 }-4x$$

$$\therefore \quad \vec { E } =-\left( \dfrac { d }{ dx } ({ 2x }^{ 2 }-4x) \right) \hat i$$
$$=(- 4x+4 ) \hat i$$

$$\therefore $$ Option (C) is the correct answer.
Question 8
1 / -0

In the adjoining figure, four capacitors are shown with the respective capacities and the $$PD$$ applied. The charge and the $$PD$$ across the $$4\ \mu\ F$$ capacitor will be:

A
$$600\ \mu C; 150\ V$$

B
$$300\ \mu C; 75\ V$$

C
$$800\ \mu C; 200\ V$$

D
$$580\ \mu C; 145\ V$$

Solution

Formula,

$$Q=CV$$

$$=\left (\dfrac{1}{20}+\dfrac{1}{8}+\dfrac{1}{12} \right )\times 300=1161\mu C$$

Charge through $$4\mu F=\dfrac{1161}{2}=580\mu C$$

Potential difference across $$4\mu F=\dfrac{580}{4}=145V$$
Question 9
1 / -0

Figure shows two capacitors connected in series and connected by a battery. The graph (B) shows the variation of potential as one moves from left and to right on the branch containing the capacitor. Then

A
$${C}_{1}={C}_{2}$$

B
$${C}_{1}< {C}_{2}$$

C
$${C}_{1}> {C}_{2}$$

D
$${C}_{1}$$ and $${C}_{2}$$ cannot be compared

Solution

According to the graph we can say that the potential difference across the capacitor $$C_1$$ is more than that across $$C_2$$

Since charge Q is same i.e,
$$Q=C_1V_1=C_2V_2$$

$$\Rightarrow \dfrac{C_1}{C_2}=\dfrac{V_2}{V_1}$$

$$\Rightarrow C_1<C_2$$ $$\left(V_1>V_2\right)$$
Question 10
1 / -0

Electrons are caused to fall through a potential difference of 1500 volt. If they were initially at rest, their final speed is

A
$$4.6 \times {10^7}m/s$$

B
$$0.726 \times {10^7}m/s$$

C
$$0.23 \times {10^2}m/s$$

D
$$5.1 \times {10^9}m/s$$

Solution

V = 1500 V
u = 0, initial K.E = 0
Energy last by electron = $$V.e = 1500 \times 1.602 \times {10^{ - 19}}$$
$$ \Rightarrow 2.403 \times {10^{ - 19}}\,\,joules$$
$$\begin{array}{l} mass\, \, of\, \, electrons\left( m \right) =9.1\times { 10^{ -31 } }kg \\ kinetic\, \, energy\, \, gained=P.E\, \, lost \\ \frac { 1 }{ 2 } m{ V^{ 2 } }=2.403\times { 10^{ -17 } } \\ { V^{ 2 } }=\frac { { 2\times 2.403\times { { 10 }^{ -17 } } } }{ { 9.1\times { { 10 }^{ -31 } } } } =\frac { { 4.806\times { { 10 }^{ +14 } } } }{ { 9.1 } } =0.528\times { 10^{ 14 } } \\ { V^{ 2 } }\sqrt { 0.528\times { { 10 }^{ 14 } } } =0.726\times { 10^{ 7 } }m/s \end{array}$$