Electrostatic Potential and Capacitance Test

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Electrostatic Potential and Capacitance Test - 24

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Question 1
1 / -0

A metal plate of thickness half the separation between the capacitor plates of capacitance C is inserted. The new capacitance is:

A
C

B
$$\frac{C}{2}$$

C
zero

D
2C

Solution

Initial capacitance $$=\frac {\varepsilon_0 A}{d}$$
Now after metal plate is inserted it behaves as two capacitor plates in series with capacitance $$\frac {\varepsilon_0 A}{(\frac {d}{a})}$$
We know when capacitance are in series
$$\frac {1}{C_{eff}}=\frac {1}{C_1}+\frac {1}{C_2}$$
$$\Rightarrow \frac {1}{C_{eff}}=\frac {d}{4\varepsilon_0 A}+\frac {d}{4\varepsilon_0 A}$$
$$\Rightarrow C_{eff}=\frac {2\varepsilon_0 A}{d.}$$
$$\therefore Now \ C_{eff}= 2C$$
Question 2
1 / -0

The equivalent capacitance between P and Q of the given figure is (the capacitance of each capacitor is$$1\ \mu F$$ ):

A
$$2\ \mu F$$

B
$$0.5\ \mu F$$

C
$$5\ \mu F$$

D
$$0.2\ \mu F$$

Solution

We know when capacitors are in series:
$$ \dfrac{1}{C_{eff}}=\dfrac{1}{C_{1}}+\dfrac{1}{C_{2}}$$
and when in parallel:
$$C_{eff}=C_{1}+C_{2}$$

So, for the given combination, between P and Q there are 2 combinations of 2 parallely connected capacitors and these combinations come in series with a single capacitor.

So, effective capacitance between P and Q may expressed as:
$$\dfrac{1}{C_{eff}}=\dfrac{1}{2C}+\dfrac{1}{C}+\dfrac{1}{2C}$$
or, $$C_{eff}=\dfrac{C}{2}$$$$=0.5 C$$
$$\therefore C_{eff}=0.5\ \mu F$$
Question 3
1 / -0

A parallel plate capacitor is first charged and then isolated , and a dielectric slab is introduced between the plates. The quantity that remains unchanged is:

A
charge $$Q$$

B
potential $$V$$

C
capacity $$C$$

D
energy $$U$$

Solution

When the capacitor is kept at a voltage, it gains charge.
Now when the system is isolated, the charge present on capacitor cannot change because of law of conservation of charge.
$$\therefore $$ Charge always remains constant in isolated systems.
Question 4
1 / -0

$$100J$$ of work is done when $$2 \mu C$$ charge is moved in an electric field between two points. The p.d. between the points is

A
$$2\times10^{-4}V$$

B
$$2\times10^{-8}V$$

C
$$2\times10^{-6}V$$

D
$$5\times10^{7}V$$

Solution

We know work done $$=q\Delta V $$
Where $$\Delta V $$ is change in potential $$V_2-V_1$$
$$100=\Delta V\times 2\times 10^{-6}$$
$$50\times 10^6=\Delta V$$
$$\therefore p\cdot d= 5\times 10^7 V$$
Question 5
1 / -0

When air is replaced by a dielectric medium of constant $$K$$, the capacity of the condenser:

A
increases $$K$$ times

B
increases $$K^{2}$$ times

C
remains unchanged

D
decreases $$K$$ times

Solution

We know, capacity: $$C=\dfrac {\varepsilon_0 A}{d}$$
Now when air is replaced by dielectric medium of constant $$k$$
$$C'=\dfrac {K\varepsilon_0 A}{d}$$
This implies that the capacitance increases $$K$$ times.
Question 6
1 / -0

$$1$$ volt is equal to

A
$${1 \ Joule}$$

B
$$\dfrac{1 \ Joule}{1 \ Coulomb}$$

C
$$\dfrac{1 \ Joule}{1 \ meter}$$

D
$$\dfrac{1 \ Newton}{1 \ Coulomb}$$

Solution

Volt is unit of Voltage which is Energy per Charge , i.e, Joule/Coulomb
Question 7
1 / -0

The work done in moving a single positive charge from infinity to a point is called ..... at that point

A
Electricity

B
Electric potential

C
Potential gradient

D
None of these

Solution

The work done is moving a single positive charge from infinity to a point is called electric potential at that point.
Work done $$=V=E.d$$
Question 8
1 / -0

Three capacitors of $$3 \mu F, 2 \mu F $$ and $$6 \mu F$$ are connected in series. When a battery of 10V is connected to this combination then charge on $$3 \mu F$$capacitor will be :

A
$$5 \mu C$$

B
$$10 \mu C$$

C
$$15 \mu C$$

D
$$20 \mu C$$

Solution

$$q=CV\ \ \ \dfrac{1}{C_{eff}}=\dfrac{1}{C_1}+\dfrac{1}{C_2}+\dfrac{1}{C_3}$$
$$\dfrac{1}{C_{ef}}=\dfrac{1}{3}+\dfrac{1}{2}+\dfrac{1}{6}$$
$$\therefore C_{ef}=1 \mu F$$
$$\therefore q=C_{ef} V = 10 \mu C$$
q is same for all capacitors
$$\therefore$$ q on any capacitor is $$10 \mu C$$
Question 9
1 / -0

Three equal condensers of capacity $$C$$ each are connected to form a triangle as shown above. The effective capacity across any side is :

A
$$3C/2$$

B
$$3C$$

C
$$C/3$$

D
$$C/2$$

Solution

Two capacitor are in series ,

$$\dfrac { 1 }{ Ceq } ={ \dfrac { 1 }{ C } }+\dfrac { 1 }{ C } \Longrightarrow Ceq=\dfrac { C }{ 2 } $$

Now this capacitor is in parallel with the C ,
$${ C }_{ e }=C+\dfrac { C }{ 2 } \\ { C }_{ e }=\dfrac { 3C }{ 2 } $$
Question 10
1 / -0

Which material sheet should be placed between the plates of a parallel plate condenser in order to increase its capacitance ?

A
mica

B
copper

C
tin

D
iron

Solution

Here copper, tin, iron all are conductor so they will decrease the capacitance. The mica sheet is a dielectric or insulator so it will increase the capacitance k times. Where k is the dielectric constant.