Electrostatic Potential and Capacitance Test

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Electrostatic Potential and Capacitance Test - 38

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Weekly Quiz Competition

Question 1
1 / -0

The net charge on a condenser is :

A
infinity

B
q/2

C
2q

D
zero

Solution

Every condenser is made with two plates. The charge on one plate is $$+Q$$ and other is $$-Q$$. Thus total charge of condenser is $$Q_t=+Q-Q=0$$
Question 2
1 / -0

$$n$$ equal capacitors are first connected in series and then in parallel. The ratio of capacitances in series and parallel arrangements will be :

A
$$n$$

B
$$n^2$$

C
$$\dfrac 1n$$

D
$$\dfrac {1}{n^2}$$

Solution

When n capacitors are connected in series, the equivalent capacitance is $$C_{series}=\dfrac{C}{n}$$
and when n capacitors are connected in parallel, the equivalent capacitance is $$C_{parallel}=nC$$
$$\dfrac{C_{series}}{C_{parallel}}=\dfrac{\dfrac{C}{n}}{nC}=\dfrac{1}{n^2}$$
Question 3
1 / -0

Three identical capacitors, each of capacitance $$C$$, are connected in series with a battery of emf $$V$$ and get fully charged. Now, the battery is removed and the capacitors are connected in parallel with positive terminals are one point and negative terminals at other point. Then, the common potential will be :

A
$$V$$

B
$$3V$$

C
$$\dfrac V3$$

D
$$0$$

Solution

When capacitors are in series ,in this case capacitance of all are same and charge on all the capacitors are also same. Hence voltage will also be same
$$V=3V'$$
$$V'=\dfrac{V}{3}$$
Now if we connect them in parallel, all are at common potential difference V/3.
Question 4
1 / -0

A slab X is placed between the two parallel isolated charged plates as shown in the figure. If $$E_p$$ and $$E_q$$ denotes the intensity of electric field at P and Q, then :

A
$$E_p$$ is reduced by the presence of X, if X is metallic

B
$$E_q$$ is increased by presence of X, if X is dielectric

C
$$E_q$$ is in the opposite sense to $$E_p$$, if X is dielectric

D
$$E_q$$ is zero, if X is metallic

Solution

There is no effect of the metal on the external electric field(thus A is incorrect) while the dielectric reduces the net electric field outside. The dielectric produces an electric field inside it due to the induced charges which is opposite to the external field. Thus $$E_Q$$ is reduced and B and C are also incorrect.(Note that $$E_P$$ is not the external field). Now as the electric field inside the conductor is zero, the field $$E_p$$ is zero if X is a metallic. Thus D is correct.
Question 5
1 / -0

Seven capacitors each of capacitance $$2\mu F$$ are to be connected in a configuration to obtain an effective capacitance of $$10/11 \mu F$$. Which of the following combination(s) shown below will archive the desired result ?

A
1

B
2

C
3

D
4

Solution

We know the equivalent capacitance for parallel combination can be found by using formula
$$C_{eq} = C_1 +C_2+C_3+....$$

And, the the equivalent capacitance for series combination can be found by using formula
$$\dfrac{1}{C_{eq}} =$$$$\dfrac{1}{C_{1}} +$$$$\dfrac{1}{C_{2}} $$+$$\dfrac{1}{C_{3}} +....$$

Option A-
Two capacitor are in series with the combination of 5 capacitor in parallel
So, $$\dfrac{1}{C_{eq}} =$$$$\dfrac{1}{2} +$$$$\dfrac{1}{2} $$+$$\dfrac{1}{(2+2+2+2+2)} $$
$$\Rightarrow C_{eq} = \dfrac{10}{11} \mu F$$

Therefore, A is correct option.
Question 6
1 / -0

A person has only two capacitors. By using them singly, in parallel or in series, he is able to obtain the capacitance $$2\mu F, 3\mu F, 6\mu F $$ and $$9\mu F$$. What are the capacitance of the capacitors?

A
$$2 \mu F$$ and $$3\mu F$$

B
$$3 \mu F$$ and $$6\mu F$$

C
$$6\mu F$$ and $$9\mu F$$

D
$$2 \mu F$$ and $$9\mu F$$

Solution

For (A), $$C^s_{eq}=\frac{2\times 3}{2+3}=\frac{6}{5} \mu F$$
$$C^p_{eq}=2+3=5 \mu F$$
For (B), $$C^s_{eq}=\frac{3\times 6}{3+6}=2 \mu F$$
$$C^p_{eq}=3+6=9 \mu F$$
For (C), $$C^s_{eq}=\frac{6\times 9}{6+9}=\frac{18}{5} \mu F$$
$$C^p_{eq}=6+9=15 \mu F$$
For (D), $$C^s_{eq}=\frac{2\times 9}{2+9}=\frac{18}{11} \mu F$$
$$C^p_{eq}=2+9=11 \mu F$$
Thus only (B) will give the capacitances which are expectable.
Question 7
1 / -0

Seven capacitors, each of capacitance $$2\mu F$$, are to be combined to obtain a capacitance of $$\dfrac{10}{11} \mu F$$. Which of the following combinations is possible ?

A
$$2$$ in parallel, $$5$$ in series connected in series

B
$$3$$ in parallel, $$4$$ in series connected in series

C
$$4$$ in parallel, $$3$$ in series connected in series

D
$$5$$ in parallel, $$2$$ in series connected in series

Solution

For A: $$C_p=2+2=4 \mu F$$ and $$C_s=(1/2+1/2+1/2+1/2+1/2)^{-1}=2/5 \mu F$$ so $$C_{eq}=\dfrac{C_pC_s}{C_p+C_s}=\dfrac{4\times 2/5}{4+2/5}=\dfrac{8}{5}\times \dfrac{5}{22}=\dfrac{4}{11} \mu F$$.

For B: $$C_p=2+2+2=6 \mu F$$ and $$C_s=(1/2+1/2+1/2+1/2)^{-1}=1/2 \mu F$$ so $$C_{eq}=\dfrac{C_pC_s}{C_p+C_s}=\dfrac{6\times 1/2}{6+1/2}=3\times \dfrac{2}{13}=\dfrac{6}{13} \mu F$$.

For C: $$C_p=2+2+2+2=8 \mu F$$ and $$C_s=(1/2+1/2+1/2)^{-1}=2/3 \mu F$$ so $$C_{eq}=\dfrac{C_pC_s}{C_p+C_s}=\dfrac{8\times 2/3}{8+2/3}=\dfrac{16}{3}\times \dfrac{3}{26}=\dfrac{8}{13} \mu F$$.

For D: $$C_p=2+2+2+2+2=10 \mu F$$ and $$C_s=(1/2+1/2)^{-1}=1 \mu F$$ so
$$C_{eq}=\dfrac{C_pC_s}{C_p+C_s}=\dfrac{10\times 1}{10+1}=\dfrac{10}{11} \mu F$$.
Question 8
1 / -0

Find the capacitance between $$P$$ and $$O$$ as shown in the figure. Each capacitor has capacitance $$C$$ :

A
$$2C$$

B
$$3C$$

C
$$8C$$

D
$$6C$$

Solution

The four capacitors with capacitance C are in parallel in between P and A.
So equivalent capacitance of parallel combination $$C_{PA}=C+C+C+C=4C$$
Similarly, $$C_{AO}=4C$$
Now $$C_{PA}$$ and $$C_{AO}$$ are in series, so equivalent capacitance of series combination between P and O is $$C_{PO}=\dfrac{C_{PA}C_{AO}}{C_{PA}+C_{AO}}=\dfrac{4C\times 4C}{4C+4C}=2C$$
Question 9
1 / -0

Arrangement of parallel plate capacitors is shown in the figure. All the plates are of same area a and placed at same distance d apart. If ends x and w are joined together through a conductor, net capacitance at terminals P and Q is given by :

A
$$\displaystyle\frac{3\in_0 A}{d}$$

B
$$\displaystyle\frac{2\in_0 A}{3d}$$

C
$$\displaystyle\frac{3\in_0A}{2d}$$

D
$$\displaystyle\frac{\in_0A}{3d}$$

Solution

Here the four plates will make $$(4-1)=3$$ capacitors with capacitance of each is $$C=\dfrac{A\epsilon_0}{d}$$
For calculating net capacitance in between P and Q we draw a equivalent circuit in figure below.
Net capacitance is $$C_N=\dfrac{CC}{C+C}+C=\dfrac{3}{2}C=\dfrac{3A\epsilon_0}{2d}$$
Question 10
1 / -0

Two identical parallel plate capacitors are connected in series and then joined in series with a battery of $$100V$$. A slab of dielectric constant $$K=3$$ is inserted between the plates of the first capacitor. Then, the potential difference across the capacitors will be respectively,

A
$$25V, 75V$$

B
$$75V, 25V$$

C
$$20V, 780

D
$$50V, 50V$$

Solution

After inserting a dielectric plate $$C_1=3C and C_2=C$$
$${ V }_{ 1 }=\cfrac { C\times V }{ 3C+C } =\cfrac{C\times 100}{4C}=25V,\quad { V }_{ 2 }=100-25=75V$$