Electrostatic Potential and Capacitance Test

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Electrostatic Potential and Capacitance Test - 33

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Question 1
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A parallel plate air capacitor is connected to a battery. The quantities charge, voltage, electric filed and energy associated with this capacitor are gives by $$Q_{0},V_{0},E_{0}$$ and $$U_{0}$$ respectively. A dielectric slab is now introduced to fill the space between the plates with the battery still in connection. The corresponding quantities are now given by $$Q$$, $$V$$, $$E$$ and $$U$$ are related to the previous one as:

A
$$Q > Q_{0}$$

B
$$V > V_{0}$$

C
$$E > E_{0}$$

D
$$U > U_{0}$$

Solution

$$Q_0=C_0 V_0$$
$$U_0=\dfrac {1}{2}C_0 V_{)}^{2}$$
$$E_0=\dfrac {V_0}{(d)}$$ (d is distance between plates).
when dielectric is added.
$$c=kC_0$$
As battery is still connected $$U_0$$ remains same.
$$V=U_0$$
$$U=\dfrac {1}{2}(kC_0)V_{0}^{2}$$
$$U-k\dfrac {1}{2} C_0 V_{0}^{2}$$
$$U=k V_0$$
$$Q=k C_0\cdot V_0.$$
$$\therefore Q=k Q_0$$
$$E=\dfrac {U}{d}$$
$$E=\dfrac {U_0}{d}$$
$$E=E_0.$$
Question 2
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The electric potential at a point in free space due to a charge Q coulomb is $$Q\times 10^{11}$$ volts. The electric field at that point is -

A
$$4\pi \epsilon _{0}\times 10^{20}volt/m$$

B
$$12\pi \epsilon _{0}\times 10^{22}volt/m$$

C
$$4\pi \epsilon _{0}\times 10^{22}volt/m$$

D
$$12\pi \epsilon _{0}\times 10^{20}volt/m$$

Solution

Since
$$V=\frac{Q}{4\pi \varepsilon _{0}r}$$ and $$E=\frac{Q}{4\pi \varepsilon _{0}r^{2}}$$;
Given $$V = Q \times 10^{11} = Q 4\pi \varepsilon _{0}r$$
$$\Rightarrow r = \frac{1}{4\pi \varepsilon _{0}\times 10^{11}}$$
i.e., $$E=\frac{V}{r}$$
$$\Rightarrow E=\frac{QV}{r}=Q \times 10^{11} \times 4\pi \varepsilon _{0}\times 10^{11}$$
$$=4\pi \varepsilon _{0}Q\times10^{22}volt \:m^{-1}$$
Question 3
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Between the plates of a parallel plate capacitor of capacity C, two parallel plates of the same material and area same as the plate of the original capacitor, are placed. If the thickness of these plates is equal to $$\left\{\dfrac {1}{5}\right\}^{th}$$ of the distance between the plates of the original capacitor, then the capacity of the new capacitor is :

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

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

C
$$\dfrac {3}{10}C$$

D
$$\left (\dfrac {10}{3} \right )C$$

Solution

Initial capacitance
$$C=\dfrac {\varepsilon_0 A}{d}$$

Now assume the plates are arranged as follows.

The above arrangement acts as three capacitors kept in series.

$$C_1=\dfrac {\varepsilon_0 A}{x_1}, C_2=\dfrac {\varepsilon_0 A}{x_2}, C_3=\dfrac {\varepsilon_0 A}{x_3}$$

$$\dfrac {1}{C_{eff}}=\dfrac {1}{C_1}+\dfrac {1}{C_2}+\dfrac {1}{C_3}$$

$$\Rightarrow \dfrac {1}{C_{eff}}=\dfrac {x_1+x_2+x_3}{\varepsilon_0 A}$$

$$\Rightarrow C_{eff}=\dfrac {\varepsilon_0 A}{x_1+x_2+x_3}$$

$$x_1+x_2+x_3=\dfrac {3d}{5}$$

$$\therefore C_{eff}=\dfrac {5\varepsilon_0 A}{3d}=\dfrac {5C}{3.}$$
Question 4
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The capacitance of a parallel plate condenser is $$C_{1}$$(fig. a). A dielectric of dielectric constant ‘K’ is inserted as shown in figure ‘b’ and ‘c’. If $$C_{2}$$ and $$C_{3}$$ are the capacitances in figures ‘b’ and ‘c’ then :

A
Both C$$_{2}$$ and C$$_{3}$$>C$$_{1}$$

B
C$$_{3}$$>C$$_{1}$$ and C$$_{2}$$>C$$_{1}$$

C
Both C$$_{2}$$ and C$$_{3}$$< C$$_{1}$$

D
C$$_{1}$$=C$$_{2}$$=C$$_{3}$$

Solution

We know that capacitance is given by
$$C_1=\dfrac {\varepsilon_0 A}{d.}$$
b) In $$C_2\dfrac {k \varepsilon_0 A 2}{d} and \dfrac {\varepsilon_o A 2}{d}$$ are in series
$$\Rightarrow \dfrac {1}{C_2}=\dfrac {d}{k\varepsilon A 2}+\dfrac

{d}{2\varepsilon_0 A}=\dfrac {d}{2\varepsilon_0 A}(\dfrac {1}{k}+1)$$
$$\Rightarrow C_2=\dfrac {\varepsilon_0 A}{d}(\dfrac {2k}{1+k})$$ we know k>1
$$\therefore \dfrac {2k}{1+k}>1.$$
$$\therefore C_2>C_1.$$

c) in $$C_3\dfrac {k\varepsilon_0 A}{2d} and \dfrac {k\varepsilon_o A}{d}$$ are in parallel
$$\therefore C_3=\dfrac {\varepsilon_0 A}{d}(\dfrac {k}{2}+1)$$
$$\therefore C_3>C_1.$$
Question 5
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A parallel plate capacitor with plates separated by air acquires 1 $$\mu $$C of charge when connected to a battery of 500V. The plates still connected to the battery are then immersed in benzene ($$k=2.25$$). Then a charge flows from the battery is :

A
$$1.25 \mu $$ C

B
$$2.28 \mu $$ C

C
$$1/4 \mu $$ C

D
$$4.56 \mu $$ C

Solution

Initial charge $$Q_1=1\mu c$$
$$V_1=500V.$$
As the plates are still connected voltage is constant.
$$\therefore $$ when plates are dipped in benzene capacitance increases by $$k=(\dfrac {k\varepsilon_0 A}{d})$$
$$\therefore $$ As voltage is constant Q also increases by k.
$$\therefore Q_2=kQ_1$$
charge flown from battery$$=kQ_1-Q_1$$
$$1\mu c(2\cdot 25-1)$$
$$1\times 10^{-6}(1\cdot 25)$$
$$=1\cdot 25\mu c.$$
Question 6
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Three capacitors each of capacity $$4 \mu F$$ are to be connected in such a way that we get the effective capacitance of $$6\mu F$$. This can be done by:

A
connecting all of them in series

B
connecting them in parallel

C
connecting two in series and one in parallel

D
connecting two in parallel and one in series

Solution

To get equivalent capacitance $$6\mu F$$ two $$4\mu F$$ capacitors are to be connected in series and third one is connected in parallel.
From figure equivalent capacitance
$${ C }_{ eq }=\cfrac { 4\times 4 }{ 4+4 } +4=2+4=6\mu F$$
Question 7
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Two short dipoles are situated parallel to each other separated by a distance x. The force of interaction is

A
A force of attraction of $$\frac{2Kp_1p_2}{x^4}$$

B
A force of repulsion of $$\frac{2Kp_1p_2}{x^4}$$

C
A force of attraction of $$\frac{3Kp_1p_2}{x^4}$$

D
A force of repulsion of $$\frac{3Kp_1p_2}{x^4}$$

Solution

Electrostatic field due to dipole 1
$$E=\frac{kp_1}{x^3}$$
now second dipole is mode of two charges say one is at a distance x of other at (x + d), $$d\rightarrow $$ length of dipole
$$\because $$ force of charge 1, $$F_1=\frac{kp_1q}{x^3}$$
force on charge 2, $$F_2=\frac{-kp_1q}{(x+d)^3}$$
both will be in opposite direction, as one is positive charge and other is negative net force
$$f = f_1+f_2$$
=$$kp_1q\left ( \frac{1}{x^3}-\frac{1}{(x+d)^3} \right )$$
=$$kp_1q\left ( \frac{x^3+3x^2d+3d^2x+d^3-x^3}{x^3(x+d)^3} \right )$$
=$$k{ p }_{ 1 }q\frac { \left( 3\left( \frac { d }{ x } \right) +3\left( \frac { d }{ x } \right) ^{ 2 }+\left( \frac { d }{ x } \right) ^{ s } \right) }{ (d+x)^{ 3 } } $$
as d < < x ignoring higher power of d/x and d as compound to r
=$$\frac{kp_1q3\left ( \frac{d}{x} \right )}{x^3}$$
=$$\frac{3kp_1(qd)}{x^4}$$
=$$\frac{3kp_1p_2}{x^4}$$
Question 8
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Positive charge flow from a body at ________to a body at _______. Fill in the blanks.

A
higher potential, lower potential

B
lower potential, higher potential

C
higher charge, lower charge

D
higher force, lower force

Solution
Question 9
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Point charges $$q_1 = +1 \: \mu C$$ and $$q_2$$ whose magnitude is $$64=27\:\mu C$$ are fixed 5 m apart along a vertical line with $$q_1$$ being at lower position. These two charges together are able to hold an oil drop of mass $$1 \mu g $$ and charge Q stationary when it is 3 m away from $$q_1$$ and 4 m away from $$q_2$$. The sign of the charge $$q_2$$ and the value of Q are respectively:

A
$$q_2 \: is \: positive, Q = 6.25 \: pC$$

B
$$q_2 \: is \: positive, Q = 6 \: pC.$$

C
$$q_2 \: is \: negative, Q = 6.25 \: pC$$

D
$$q_2 \: is \: negative, Q = 6 \: pC.$$
Question 10
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The potential difference between points A and B, in a section of a circuit shown, is

A
5 Volt

B
1 Volt

C
10 Volt

D
-16Volt

Solution

assuming the pot at A to be zero
$$V_A-6-3-4+2-5=V_B$$
$$\Rightarrow V_B-V_A=-16$$