Electrostatic Potential and Capacitance Test

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Electrostatic Potential and Capacitance Test - 39

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Question 1
1 / -0

The separation between the plates of $$C_1$$ of Fig is doubled and a dielectric of strength 6 is added in between the plates of $$C_2$$. Find the the change in potential drop across $$C_1$$ and $$C_2$$:

A
$$\Delta V_1=\dfrac{11C_2V_0C_1}{(C_1+12C_2)(C_1+C_2)}(increases)$$

$$\Delta V_2=\dfrac{11C_1C_2V_0}{(C_1+C_2)(C_1+12C_2)}(decreases)$$

B
$$\Delta V_1=\dfrac{11C_2V_0C_1}{(C_1+12C_2)(C_1+C_2)}(decreases)$$

$$\Delta V_2=\dfrac{11C_1C_2V_0}{(C_1+C_2)(C_1+12C_2)}(increases)$$

C
$$\Delta V_1=\dfrac{22C_2V_0C_1}{(C_1+12C_2)(C_1+C_2)}(increases)$$

$$\Delta V_2=\dfrac{22C_1C_2V_0}{(C_1+C_2)(C_1+12C_2)}(decreases)$$

D
$$\Delta V_1=\dfrac{22C_2V_0C_1}{(C_1+12C_2)(C_1+C_2)}(decreases)$$

$$\Delta V_2=\dfrac{22C_1C_2V_0}{(C_1+C_2)(C_1+12C_2)}(increases)$$

Solution

$$C=\epsilon A/d$$ $$d_1=2d_0$$ $$\epsilon_2=6\epsilon_0$$
$$C'_1=\dfrac{C_1}{2} and C'_2=6C_2$$
from circuit of Fig 20.60(a)
$$V_1=\dfrac{C_2V_0}{C_1+C_2}\ V_2=\dfrac{C_1V_0}{C_1+C_2}$$
From circuit of Fig 20.60 (b)
$$V'_1=\dfrac{6C_2V_0}{\dfrac{C_1}{2}+6C_2}=\dfrac{12C_2V_0}{C_1+12C_1};$$
$$V'_2=\dfrac{\dfrac{C_1}{2}V_0}{\dfrac{C_1}{2}+6C_2}=\dfrac{C_1V_0}{C_1+12C_2}$$
$$\Delta V_1=\dfrac{12C_2V_0}{C_1+12C_2}-\dfrac{C_2V_0}{C_1+C_2}$$
$$=C_2V_0\left [ \dfrac{12(C_1+C_2)-(C_1+12C_2)}{(C_1+12C_2)(C_1+C_2)} \right ]$$
$$=\dfrac{11C_2V_0C_1}{(C_1+12C_2)(C_1+C_2)}(increases)$$
$$\Delta V_2=\dfrac{C_1V_0}{C_1+C_2}-\dfrac{C_1V_0}{C_1+12C_2}$$
$$=C_1V_0\left ( \dfrac{C_1+12C_2-(C_1+C_2)}{(C_1+C_2)(C_1+12C_2)} \right )$$
$$=\dfrac{11C_1C_2V_0}{(C_1+C_2)(C_1+12C_2)}(decreases)$$
Question 2
1 / -0

An electron is taken from point $$A$$ to point $$B$$ along the path $$AB$$ in a uniform electric field of intensity $$\displaystyle E = 10 Vm^{-1}$$. Side $$AB = 5 m$$, and side $$BC = 3 m$$. Then, the amount of work done on the electron by us is :

A
50 eV

B
40 eV

C
-50 eV

D
-40 eV

Solution

$$\displaystyle W_{AB} = W_{AC}+W_{CB}$$
$$\displaystyle W_{CB}$$ should be zero, because in moving from $$C$$ to $$B$$, we always move perpendicular to field. Hence, force applied by field and displacement will be at $$90^{\circ}$$.
so work done in BC will be 0
$$\displaystyle W_{AC} = -e(V_C-V_A)$$
$$\displaystyle V_C-V_A = -E \times AC = -10 \times 4 = -40$$
$$\displaystyle \therefore W_{AB} = 40e J=40 eV$$
Question 3
1 / -0

A parallel plate capacitor is charged to a certain voltage. Now, if the dielectric material (with dielectric constant k) is removed then the

A
capacitance increases by a factor of k

B
electric field reduces by a factor k

C
voltage across the capacitor decreases by a factor k

D
none of these

Solution

As the capacitor is charged by using cell so potential as well as filed between the plates become constant.
For removing dielectric the capacitance becomes $$C/k$$. Thus capacitance decreases by a factor of k.
Question 4
1 / -0

Two parallel metal plates having charges +Q and -Q face each other at a certain distance between them. If the plates are now dipped in kerosene oil tank, the electric field between the plates will :

A
remain same

B
become zero

C
increase

D
decrease

Solution

The electric field in between two parallel metal plate is $$E=\dfrac{\sigma}{\epsilon}=\dfrac{Q}{A\epsilon}$$ where $$A=$$ area of plates and $$\epsilon=$$ permeability of medium in between the plates.
The permeability of kerosene is more that of air. As $$\epsilon$$ increases, $$E$$ decreases.
Question 5
1 / -0

A foil of aluminium of negligible thickness is inserted in between the space of a parallel plate condenser. If the foil is electrically insulated, the capacity of the condenser will :

A
increase

B
decrease

C
remain unchanged

D
become zero

Solution

The capacity of condenser before inserting foil is $$C=\dfrac{A\epsilon_0}{d}$$ where A be the area of plate and $$d$$ be the separation between plates.
After inserting foil the there will be two capacitors in series with capacitance 2C as distance is halved and the series combination of the two will give equivalent capacitance of C, hence, capacity will remain same.
Question 6
1 / -0

If in a parallel plate capacitor, which is connected to a battery, we fill dielectrics in whole space of its plates, then which of the following increases?

A
Q and V

B
V and E

C
E and C

D
Q and C

Solution

Since battery remains connected so P.D. between the plates is constant. But as we introduce the dielectric the capacitance increases and hence charge increases.
Question 7
1 / -0

A parallel plate condenser with oil between the plates (dielectric constant of oil $$K=2)$$ has a capacitance C. If the oil is removed, then capacitance of the capacitor becomes :

A
$$\sqrt 2C$$

B
$$2C$$

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

D
$$\dfrac {C}{2}$$

Solution

When oil is present, the capacitance $$\displaystyle C=\frac{AK\epsilon_0}{d}=\frac{2A\epsilon_0}{d}$$ where $$A=$$ area of plates and $$d=$$ separation of plates and dielectric constant $$K=2$$
When oil is removed, the capacitance becomes $$\displaystyle C'=\frac{A\epsilon_0}{d}=\frac{C}{2}$$
Question 8
1 / -0

A parallel plate capacitor with air between the plates is charged to a potential difference of $$500 V$$ and then insulated. A plastic plate is inserted between the plates filling the whole gap. The potential difference between the plates now becomes $$75V$$. The dielectric constant of plastic is :

A
$$10/3$$

B
$$5$$

C
$$20/3$$

D
$$10$$

Solution

If $$C$$ be the air filled capacitance , dielectric filled capacitance becomes $$C'=KC$$ where K is the dielectric constant.
Initial charge , $$q_i=CV_i=500 C$$ and final charge $$q_f=C'V_f=75KC$$
As the system is isolated so charge will be constant.
Thus, $$q_i=q_f $$
$$\Rightarrow 500C=75 KC \Rightarrow K=\dfrac{500}{75}=20/3$$
Question 9
1 / -0

Three capacitors each of capacity $$4\mu F$$ are to be connected in such a way that the effective capacitance is $$6\mu F$$. This can be done by :

A
connecting two in parallel and one in series

B
connecting all of them in series

C
connecting them in parallel

D
connecting two in series and one in parallel

Solution

A) $$C_{eff}=\dfrac{(4+4)4}{(4+4)+4}=\dfrac{32}{12}=\dfrac{8}{3} \mu F$$

B) $$C_{eff}=\dfrac{C}{n}=\dfrac{4}{3} \mu F$$

C) $$C_{eff}=nC=3(4)=12 \mu F$$

D) $$C_{eff}=\dfrac{4\times 4}{4+4}+4=2+4=6 \mu F$$
Question 10
1 / -0

Two capacitors of capacitance C are connected in series. If one of them is filled with dielectric substance k, what is the effective capacitance?

A
$$\frac {kC}{(1+k)}$$

B
C(k+1)

C
$$\frac {2kC}{1+k}$$

D
none of these

Solution

When capacitance filled with dielectric , the capacitance becomes $$C'=kC$$
As they are in series so effective capacitance is $$\displaystyle C_{ef}=\frac{CC'}{C+C'}=\frac{C(kC)}{C+kC}=\frac{kC}{1+k}$$

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Electrostatic Potential and Capacitance Test - 39

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Question 1

$$\Delta V_1=\dfrac{11C_2V_0C_1}{(C_1+12C_2)(C_1+C_2)}(increases)$$$$\Delta V_2=\dfrac{11C_1C_2V_0}{(C_1+C_2)(C_1+12C_2)}(decreases)$$

$$\Delta V_1=\dfrac{11C_2V_0C_1}{(C_1+12C_2)(C_1+C_2)}(decreases)$$$$\Delta V_2=\dfrac{11C_1C_2V_0}{(C_1+C_2)(C_1+12C_2)}(increases)$$

$$\Delta V_1=\dfrac{22C_2V_0C_1}{(C_1+12C_2)(C_1+C_2)}(increases)$$$$\Delta V_2=\dfrac{22C_1C_2V_0}{(C_1+C_2)(C_1+12C_2)}(decreases)$$

$$\Delta V_1=\dfrac{22C_2V_0C_1}{(C_1+12C_2)(C_1+C_2)}(decreases)$$$$\Delta V_2=\dfrac{22C_1C_2V_0}{(C_1+C_2)(C_1+12C_2)}(increases)$$

Solution

Question 2

50 eV

40 eV

-50 eV

-40 eV

Solution

Question 3

capacitance increases by a factor of k

electric field reduces by a factor k

voltage across the capacitor decreases by a factor k

none of these

Solution

Question 4

remain same

become zero

increase

decrease

Solution

Question 5

increase

decrease

remain unchanged

become zero

Solution

Question 6

Q and V

V and E

E and C

Q and C

Solution

Question 7

$$\sqrt 2C$$

$$2C$$

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

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

Solution

Question 8

$$10/3$$

$$5$$

$$20/3$$

$$10$$

Solution

Question 9

connecting two in parallel and one in series

connecting all of them in series

connecting them in parallel

connecting two in series and one in parallel

Solution

Question 10

$$\frac {kC}{(1+k)}$$

C(k+1)

$$\frac {2kC}{1+k}$$

none of these

Solution

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$$\Delta V_1=\dfrac{11C_2V_0C_1}{(C_1+12C_2)(C_1+C_2)}(increases)$$

$$\Delta V_2=\dfrac{11C_1C_2V_0}{(C_1+C_2)(C_1+12C_2)}(decreases)$$

$$\Delta V_1=\dfrac{11C_2V_0C_1}{(C_1+12C_2)(C_1+C_2)}(decreases)$$

$$\Delta V_2=\dfrac{11C_1C_2V_0}{(C_1+C_2)(C_1+12C_2)}(increases)$$

$$\Delta V_1=\dfrac{22C_2V_0C_1}{(C_1+12C_2)(C_1+C_2)}(increases)$$

$$\Delta V_2=\dfrac{22C_1C_2V_0}{(C_1+C_2)(C_1+12C_2)}(decreases)$$

$$\Delta V_1=\dfrac{22C_2V_0C_1}{(C_1+12C_2)(C_1+C_2)}(decreases)$$

$$\Delta V_2=\dfrac{22C_1C_2V_0}{(C_1+C_2)(C_1+12C_2)}(increases)$$