Electrostatic Potential and Capacitance Test

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Electrostatic Potential and Capacitance Test - 43

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Question 1
1 / -0

Among identical spheres A and B having charges $$-15\ C$$ and $$-16\ C$$:

A
$$-15\ C$$ is at higher potential

B
$$-16\ C$$ is at higher potential

C
both are at equal potential

D
no such comparison can be made

Solution

Potential at surface of a sphere of radius r is $$V=kQ/r$$
As identical sphere so r is same for both, Thus V will depend on charge Q.
So when Q is more , potential will be more.
Since $$-15\ C$$ is more than $$-16\ C$$, so $$-15\ C$$ sphere will have higher potential.
Question 2
1 / -0

Three condenser of capacitance $$C(\mu F)$$ are connected in parallel to which a condenser of capacitance $$C$$ is connected in series. Effective capacitance is $$3.75$$, then capacity of each condenser is

A
$$4\mu F$$

B
$$5\mu F$$

C
$$6\mu F$$

D
$$8\mu F$$

Solution

The effective capacitance of three condenser connected in parallel$$=3C$$.
When $$3C$$ is connected in series to $$C$$
$$C_{Result}=\displaystyle\frac{3C\times C}{3C+C}=3.75$$
$$\Rightarrow C=5\mu F$$.
Question 3
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A parallel plate capacitor with a dielectric slab of dielectric constant $$3$$, filling the space between the plates, is charged to a potential $$ V$$. The battery is then disconnected and the dielectric slab is withdrawn. It is then replaced by another dielectric slab of dielectric constant $$2$$. If the energies stored in the capacitor before and after the dielectric slab is changed are $$\displaystyle { E }_{ 1 }$$ and $$\displaystyle { E }_{ 2 }$$, then $$\displaystyle { E }_{ 1 }/{ E }_{ 2 }$$ is:

A
$$\displaystyle \frac { 4 }{ 9 } $$

B
$$\displaystyle \frac { 2 }{ 3 } $$

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

D
$$\displaystyle \frac { 9 }{ 5 } $$

Solution

Let the charge stored by capacitor with dielectric constant $$3$$ be $$Q$$.
Thus energy stored be $$\dfrac{Q^2}{2C_1}$$

Since the charge remains the same after changing the dielectric, the new energy stored=$$\dfrac{Q^2}{2C_2}$$

$$\implies \dfrac{E_1}{E_2}=\dfrac{C_2}{C_1}=\dfrac{K_2}{K_1}=\dfrac{2}{3}$$

$$(C=\dfrac{KA\epsilon_0}{d})$$
Question 4
1 / -0

In a capacitive circuit, dielectrics materials are placed between the plates of capacitors to

A
Speed the current flow

B
Slow the current flow

C
Reduce change leakage from the capacitor

D
increase change leakage from the capacitor

E
Increase capacitance of the capacitor

Solution

When a dielectric material with dielectric constant K is placed in between the plates of a capacitor, the capacitance will be increased by factor K. Thus, the option E will be correct.
Question 5
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Two identical capacitors are arranged in a circuit. What is the ratio of the equivalent capacitance of the circuit when the capacitors are in series to that when they are in parallel?

A
$$\dfrac{1}{4}$$

B
$$\dfrac{1}{2}$$

C
1

D
2

E
4

Solution

When the two identical capacitors are in series, the equivalent capacitance, $$C_s=\dfrac{CC}{C+C}=C/2$$
When the two identical capacitors are in parallel, the equivalent capacitance, $$C_p={C+C}=2C$$
Thus, $$\dfrac{C_s}{C_p}=\dfrac{C/2}{2C}=1/4$$
Question 6
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A dielectric is inserted into a capacitor while keeping the charge constant. Identify what happens to the potential difference and the stored energy?

A
The potential difference decreases and the stored energy increases

B
Both the potential difference and the stored energy increases

C
The potential difference increases and the stored energy decreases

D
Both the potential difference and the stored energy decrease

E
Both the potential difference and the stored energy remain the same

Solution

The capacitance of the capacitor increases due to the insertion of the dielectric, $$C' = kC$$ ; where, $$k>1$$
Initial potential difference on the capacitor $$ V = \dfrac{Q}{C}$$
New potential difference $$V' = \dfrac{Q}{C'} = \dfrac{Q}{kC} =\dfrac{V}{k}$$
Thus potential difference decreases.
Also initial energy stored in the capacitor $$E = \dfrac{Q^2}{2C}$$
New energy stored $$E' = \dfrac{Q^2}{2 C'} = \dfrac{Q^2}{2 (kC)} = \dfrac{E}{k}$$
Thus energy stored in the capacitor also decreases.
Question 7
1 / -0

Above, three arrangements of identical capacitors are shown. What is the correct order of total capacitance values for the arrangements, greatest first?

A
1,2,3

B
3,2,1

C
3,1,2

D
2,3,1

E
2,1,3

Solution

Let the capacitance of each capacitor be $$C$$.
Thus capacitance in case 1 $$C_1 = C$$
Case 2 : Equivalent capacitance in series combination $$C_2 = \dfrac{C\times C}{C+C} = \dfrac{C}{2}$$

Case 3 ; Equivalent capacitance in parallel combination $$C_3 = C+C = 2C$$
$$\implies$$ $$C_3 > C_1 > C_2$$
Hence correct answer is option C.
Question 8
1 / -0

How much power is used in moving a $$0.03C$$ charge from point A at a potential of $$40V$$ to point B at $$60V$$, if this takes 2 seconds?

A
$$0.30W$$

B
$$0.15W$$

C
$$0.6W$$

D
$$0.012W$$

Solution

Given : $$q = 0.03$$ C $$V_A = 40$$ V $$V_B = 60$$ V $$t = 2$$ s
Work done in moving a charge from A to B $$W = q(V_B - V_A) = 0.03 \times (60-40) = 0.6$$ J
$$\therefore$$ Power used $$P = \dfrac{W}{t} =\dfrac{0.6}{2} = 0.3$$ $$W$$
Question 9
1 / -0

Calculate the change in potential energy of a particle of charge $$+q$$ that is brought from a distance of $$3r$$ to a distance of $$2r$$ in the electric field of charge $$-q$$?

A
$${ k{ q }^{ 2 } }/{ r }$$

B
$${ -k{ q }^{ 2 } }/{ 6r }$$

C
$${ k{ q }^{ 2 } }/{ 4{ r }^{ 2 } }$$

D
$${ -k{ q }^{ 2 } }/{ 4{ r }^{ 2 } }$$

E
$${ k{ q }^{ 2 } }/{ { r }^{ 2 } }$$

Solution

Given : $$PB = 2r$$ $$PA = 3r$$
Potential at point A $$V_A = \dfrac{k q}{PA} = \dfrac{kq}{3r}$$

Potential at point B $$V_B = \dfrac{k q}{PB} = \dfrac{kq}{2r}$$
Change in potential $$\Delta V = V_B - V_A = \dfrac{kq}{2r} - \dfrac{kq}{3r} = \dfrac{kq}{6r}$$
Thus change in potential energy by moving a charge $$-q$$ from A to B $$W = -q (\Delta V) = \dfrac{-kq^2}{6r}$$
Question 10
1 / -0

What is the change in potential energy of a particle of charge +q that is brought from a distance of 3r to a distance of 2r by a particle of charge q?

A
$$kq^2/r$$

B
$$-kq^2/6r$$

C
$$kq^2/r^2$$

D
$$-kq^2//4r^2$$

E
$$8kq^2/r^2$$

Solution

As the $$+q$$ is brought near by charge $$q$$ so there is an attraction force it means q is a negative charge, equal to $$-q$$.
The electric potential energy of a system of two point charges ( $$+q$$ , $$-q$$) sepatrated by a distance $$3r$$ is given by ,
$$U=k\dfrac{q\times -q}{3r}$$ ,
when the separation is reduced to $$2r$$ then potential energy will become ,
$$U'=k\dfrac{q\times- q}{2r}$$
therefore change in potential energy is
$$\Delta U=U'-U$$
or $$\Delta U=-k\dfrac{q\times q}{2r}+k\dfrac{q\times q}{3r}$$
or $$\Delta U=kq^{2}\left(1/3r-1/2r\right)=-kq^{2}/6r$$

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Electrostatic Potential and Capacitance Test - 43

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Electrostatic Potential and Capacitance Test - 43

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Question 1

$$-15\ C$$ is at higher potential

$$-16\ C$$ is at higher potential

both are at equal potential

no such comparison can be made

Solution

Question 2

$$4\mu F$$

$$5\mu F$$

$$6\mu F$$

$$8\mu F$$

Solution

Question 3

$$\displaystyle \frac { 4 }{ 9 } $$

$$\displaystyle \frac { 2 }{ 3 } $$

$$\displaystyle \frac { 3 }{ 2 } $$

$$\displaystyle \frac { 9 }{ 5 } $$

Solution

Question 4

Speed the current flow

Slow the current flow

Reduce change leakage from the capacitor

increase change leakage from the capacitor

Increase capacitance of the capacitor

Solution

Question 5

$$\dfrac{1}{4}$$

$$\dfrac{1}{2}$$

1

2

4

Solution

Question 6

The potential difference decreases and the stored energy increases

Both the potential difference and the stored energy increases

The potential difference increases and the stored energy decreases

Both the potential difference and the stored energy decrease

Both the potential difference and the stored energy remain the same

Solution

Question 7

1,2,3

3,2,1

3,1,2

2,3,1

2,1,3

Solution

Question 8

$$0.30W$$

$$0.15W$$

$$0.6W$$

$$0.012W$$

Solution

Question 9

$${ k{ q }^{ 2 } }/{ r }$$

$${ -k{ q }^{ 2 } }/{ 6r }$$

$${ k{ q }^{ 2 } }/{ 4{ r }^{ 2 } }$$

$${ -k{ q }^{ 2 } }/{ 4{ r }^{ 2 } }$$

$${ k{ q }^{ 2 } }/{ { r }^{ 2 } }$$

Solution

Question 10

$$kq^2/r$$

$$-kq^2/6r$$

$$kq^2/r^2$$

$$-kq^2//4r^2$$

$$8kq^2/r^2$$

Solution

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