Electrostatic Potential and Capacitance Test

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Electrostatic Potential and Capacitance Test - 40

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Weekly Quiz Competition

Question 1
1 / -0

A positive point charge q is carried from a point B to a point A in the electric field of a point charge $$+$$Q held at O. If the permittivity of free space is $$\epsilon_0$$ and OA=a and OB=b, the work done in the process is given by :

A
$$\dfrac {qQ}{4\pi \epsilon_0}\left (\dfrac {1}{a}+\dfrac {1}{b}\right )$$

B
$$\dfrac {qQ}{4\pi \epsilon_0}\left (\dfrac {1}{a}-\dfrac {1}{b}\right )$$

C
$$\dfrac {qQ}{4\pi \epsilon_0}\left (\dfrac {1}{a^2}-\dfrac {1}{b^2}\right )$$

D
$$\dfrac {qQ}{4\pi \epsilon_0}\left (\dfrac {1}{a^2}+\dfrac {1}{b^2}\right )$$

Solution

We know that work done $$dW=qdV$$

here , $$\displaystyle W_{B\rightarrow A}=q{V_A-V_B}=q\left[\dfrac{Q}{4\pi\epsilon_0 (OA)}-\dfrac{Q}{4\pi\epsilon_0(OB)}\right]=\dfrac{qQ}{4\pi\epsilon_0}[\dfrac{1}{a}-\dfrac{1}{b}]$$ where $$OA=a, OB=b$$
Question 2
1 / -0

Two capacitors when connected in series have a capacitance of $$3\mu F$$, and when connected in parallel have a capacitance of $$16\mu F$$. Their individual capacities are :

A
$$1\mu F, 2\mu F$$

B
$$6\mu F, 2\mu F$$

C
$$12\mu F, 4\mu F$$

D
$$3\mu F, 16\mu F$$

Solution

Let their individual capacities are $$C_1$$ and $$C_2$$.
Here, $$C_S=\dfrac{C_1C_2}{C_1+C_2}=3 ...(1)$$
and $$C_P=C_1+C_2=16 ...(2)$$
From (1) and (2), $$C_1C_2=3(16)=48 ...(3)$$
From (2), (3), $$C_1+48/C_1=16 \Rightarrow C_1^2-16C_1+48=0$$
or $$C_1^2-12C_1-4C_2+48=0$$
or $$C_1(C_1-12)-4(C_1-12)=0$$
or $$(C_1-12)(C_1-4)=0$$
$$C_1=12, 4 \mu F$$
From (3), for $$C_1=12, C_2=\dfrac{48}{12}=4 \mu F$$ and for $$C_1=4, C_2=\dfrac{48}{4}=12 \mu F$$
Question 3
1 / -0

A sheet of aluminium foil of negligible thickness is introduced between the plates of a capacitor. The capacitance of the capacitor :

A
decreases

B
remains unchanged

C
becomes infinite

D
increases
Solution
- Capacitance of parellel plate capacitor with a dielectric slab of thickness t inside is given by: $$C=\dfrac{A\epsilon_0}{(d-t)+\dfrac{t}{K}}$$ , where K is the dielectric constant.
- When aluminium sheet of negligible thickness is inserted, then thickness of slab $$t\rightarrow 0$$ and value of $$K$$ for metal (conductor) will be $$\infty$$
- Therefore $$C=\dfrac{A\epsilon_0}{d}$$
- Hence, capacitance of capacitor will remains unchanged.
- Option B is correct.
Question 4
1 / -0

From a supply of identical capacitors rated $$8 mF, 250 V$$, the minimum number of capacitors required to form a composite $$16mF, 1000V$$ is :

A
$$2$$

B
$$4$$

C
$$16$$

D
$$32$$

Solution

Let 'n' such capacitors are in series and such 'm' such branch are in parallel.
Total potential across n capacitors $$=250\times n$$
here, $$250n=1000 \Rightarrow n=4$$
also $$8/n \times m=16 \Rightarrow m=16n/8=16(4)/8=8$$
The number of capacitors$$=mn=8(4)=32$$
Question 5
1 / -0

If a slab of insulating material $$4\times 10^{-5}m$$ thick is introduced between the plates of a parallel plate capacitor, the distance between the plates has to be increased by $$3.5\times 10^{-5}m$$ to restore the capacity to original value. Then the dielectric constant of the material of slab is :

A
$$8$$

B
$$6$$

C
$$12$$

D
$$10$$

Solution

If $$K$$ be the dielectric constant, the distance increased due to introduce dielectric is $$x=t-\dfrac{t}{K}=t\left (1-\dfrac{1}{K}\right )$$
where $$t=$$ thickness of dielectric.
According to question, $$x=3.5 \times 10^{-5}$$
$$\Rightarrow t\left (1-\dfrac{1}{K}\right )=3.5 \times 10^{-5}$$

$$\Rightarrow 1-\dfrac{1}{K}=\dfrac{3.5 \times 10^{-5}}{4\times 10^{-5}}=0.875$$

$$\Rightarrow \dfrac{1}{K}=1-0.875=0.125$$

$$\Rightarrow K=8$$
Question 6
1 / -0

An air capacitor C connected to a battery of e.m.f. V acquires a charge q and energy E. The capacitor is disconnected from the battery and a dielectric slab is placed between the plates. Which of the following statements is correct?

A
V and q decrease but C and E increase

B
V remains unchange, but q, E and C increase

C
q remains unchanged, C increases, V and E decrease

D
q and C increase but V and E decrease

Solution

As the capacitor is disconnected from battery so charge q will remain uncharged.
When dielectric slab placed, the capacitance becomes $$C'=kC$$. where $$k$$ is dielectric constant. Thus the capacitance C increases k times.
As $$V=\frac{q}{C}$$, V decreases because q is constant and C increases.
As $$V=Ed$$, so E also decreases because V decreases.
Question 7
1 / -0

A capacitor becomes a perfect insulator for

A
alternating current

B
direct current

C
both (a) and (b)

D
None of these

Solution

A capacitor becomes a perfect insulator for direct current as the regular supply of current charges capacitor and the it behaves as open circuit, where as in A.C. the current being variable in sign and magnitude does not charge capacitor ever it goes through a repetitive process of charging and discharging and hence it never behaves as open circuit.
Question 8
1 / -0

The electric potential inside a conductor:

A
is zero

B
increases with distance from center

C
is constant

D
decreases with distance from center

Solution

As the electric field inside a conductor is zero so the potential at any point is constant.
Suppose $$a $$ and $$ b$$ two points inside a conductor. we know that $$E=-\dfrac{dV}{dr}$$
As $$E=0$$, $$dV=0$$ or $$V_a-V_b=0$$ or $$V_a=V_b$$
Question 9
1 / -0

An electrical charge $$2 \times 10^{-8}$$ C is placed at the point (1, 2, 4 )m. At the point (4, 2, 0) m, the electric

A
potential will be 36 V

B
field will be along y-axis

C
field will increase if the space between the points is filled with a dielectric

D
All of the above

Solution

Here, charge$$q=2\times 10^{-8} C$$ and $$r=\sqrt{(1-4)^2+(2-2)^2+(4-0)^2}=\sqrt{25}=5 m$$
and $$\hat{r}=\dfrac{\vec{r}}{r}=\dfrac{1}{5}[(1-4)\hat i+(2-2)\hat j+(4-0)\hat k]=\dfrac{1}{5}[-3\hat i+4\hat k]$$
Thus, potential $$V=\dfrac{1}{4\pi\epsilon_0} \dfrac{q}{r}=9\times 10^9 \times \dfrac{2\times 10^{-8}}{5}=36 V$$
Electric filed $$\vec{E}=\dfrac{1}{4\pi\epsilon_0} \dfrac{q}{r^2} \hat r=9\times 10^9 \times \dfrac{2\times 10^{-8}}{5^2}. \dfrac{1}{5}[-3\hat i+4\hat k]=1.44[-3\hat i+4\hat k]$$
Thus, the field will along $$x-z$$ plane.
If space between points is filled by dielectric, the filed becomes $$E'=E/k$$ where $$k=$$ dielectric constant. since, $$k>1$$ so filed will decrease.
Question 10
1 / -0

If we increase '$$d$$' of a parallel condenser to '$$2d$$' and fill wax to the whole empty space between its two plate, then capacitance increase from $$1pF$$ to $$2pF$$. What is the dielectric constant of wax?

A
$$2$$

B
$$4$$

C
$$6$$

D
$$8$$

Solution

Initially the capacitance, $$C_1=\dfrac{A\epsilon_0}{d} $$ where $$A=$$ area of plates
Final capacitance, $$C_2=\dfrac{AK\epsilon_0}{2d} $$ where $$K=$$ dielectric constant.
$$\therefore \dfrac{C_1}{C_2}=\dfrac{2}{K} \Rightarrow \dfrac{1}{2}=\dfrac{2}{K}$$

$$\therefore K=4$$

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Electrostatic Potential and Capacitance Test - 40

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Electrostatic Potential and Capacitance Test - 40

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Question 1

$$\dfrac {qQ}{4\pi \epsilon_0}\left (\dfrac {1}{a}+\dfrac {1}{b}\right )$$

$$\dfrac {qQ}{4\pi \epsilon_0}\left (\dfrac {1}{a}-\dfrac {1}{b}\right )$$

$$\dfrac {qQ}{4\pi \epsilon_0}\left (\dfrac {1}{a^2}-\dfrac {1}{b^2}\right )$$

$$\dfrac {qQ}{4\pi \epsilon_0}\left (\dfrac {1}{a^2}+\dfrac {1}{b^2}\right )$$

Solution

Question 2

$$1\mu F, 2\mu F$$

$$6\mu F, 2\mu F$$

$$12\mu F, 4\mu F$$

$$3\mu F, 16\mu F$$

Solution

Question 3

decreases

remains unchanged

becomes infinite

increases

Solution

Question 4

$$2$$

$$4$$

$$16$$

$$32$$

Solution

Question 5

$$8$$

$$6$$

$$12$$

$$10$$

Solution

Question 6

V and q decrease but C and E increase

V remains unchange, but q, E and C increase

q remains unchanged, C increases, V and E decrease

q and C increase but V and E decrease

Solution

Question 7

alternating current

direct current

both (a) and (b)

None of these

Solution

Question 8

is zero

increases with distance from center

is constant

decreases with distance from center

Solution

Question 9

potential will be 36 V

field will be along y-axis

field will increase if the space between the points is filled with a dielectric

All of the above

Solution

Question 10

$$2$$

$$4$$

$$6$$

$$8$$

Solution

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