Electrostatic Potential and Capacitance Test

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Electrostatic Potential and Capacitance Test - 28

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Question 1
1 / -0

Statement(A): Negative charges always move from a higher potential to lower potential point

Statement (B): Electric potential is vector.

A
A is true but B is false

B
B is true but A is false

C
Both A and B false

D
Both A and R are true

Solution

Positive charges move from higher to lower potential whereas negative charges move from lower to higher potential. Hence, A is false.

Electric potential does not have direction and is a scalar quantity. Hence, B is false.
Question 2
1 / -0

A parallel plate condenser is charged by connecting it to a battery. Without disconnecting the battery, the space between the plates is completely filled with a medium of dielectric constant k. Then:

A
potential becomes 1/k times

B
charge becomes k times

C
energy becomes 1/k times

D
electric intensity becomes k times

Solution

When the battery is not disconnected the potential across the capacitor must be same as voltage of battery (kirchoffs voltage law)
Now when dielectric slab is added capacitance increases by k.
we know that $$ Q= CV$$
V is constant, C increases by k $$\therefore $$
Q also increases by k.
Question 3
1 / -0

A charge of $$2\ C$$ is moved from a point $$2\ m$$ away from a charge of $$1\ C$$ to a point $$1\ m$$ away from that charge. The work done is:

A
$$10^{9}\ J$$

B
$$10^{6}\ J$$

C
$$9\times 10^{9}\ J$$

D
$$10^{10}\ J$$

Solution

Voltage due to $$1\ C$$ at a distance $$r$$ is $$\dfrac {k \times 1}{r}$$.

Total change in potential in moving $$2\ C$$ charge from $$2\ m$$ away to $$1\ m$$ away:
$$V_2-V_1=\dfrac {k \times 1}{1}-\dfrac {k \times 1}{2}$$
or, $$V_2-V_1=\dfrac{k \times 1}{2} = \Delta V$$

So, work done $$=\Delta V Q$$$$=\dfrac {k \times 1}{2}\times 2=9\times 10^9\ J$$
Question 4
1 / -0

A parallel plate capacitor, filled with a material of dielectric constant $$K$$, is charged to a certain voltage and is isolated. The dielectric material is removed. Then:
(a) capacitance decreases by a factor $$K$$
(b) electric field reduces by a factor $$K$$
(c) voltage across the capacitor increases by a factor $$K$$
(d) charge stored in the capacitor increases by a factor $$K$$

A
(a) and (b) are true

B
(a) and (c) are true

C
(b) and (c) are true

D
(b) and (d) are true

Solution

Initial capacitance $$c= \dfrac {k\varepsilon_0 A}{d}$$
When the system is isolated, charge ($$Q$$) present on it is constant.
Now, when dielectric is removed $$C=\dfrac {\varepsilon_0 A}{d}$$
$$\Rightarrow$$ $$C$$ decreases by a factor of $$K$$.
We know that $$Q=C \times V$$
As $$C$$ decreases by factor of $$K$$ and $$Q$$ remains constant, $$V$$ increases by a factor of $$K$$.
Question 5
1 / -0

Twenty seven identical mercury drops each charged to $$10V$$, are allowed to form a big drop. The potential of the big drop is

A
$$90 V$$

B
$$9 V$$

C
$$900 V$$

D
$$270 V$$

Solution

Voltage of each mercury drop $$= 10 V$$
$$\Rightarrow \dfrac {kq}{r}= 10 V$$
$$\Rightarrow q= \dfrac {10r}{k} $$
$$total \ charge =\dfrac {27\times 10r}{k}$$
Equating volumes to find radius of bigger charge,

$$\dfrac {4}{3}\pi (R^3)=27\cdot \dfrac {4}{3}\pi r^3$$

$$R=3r$$
final voltage$$ = \dfrac {kq}{R}=\dfrac {k \times 27\times 10r}{k\times 3r}$$
$$=90 V$$
Question 6
1 / -0

(1): The dielectric medium between the plates of a parallel plate capacitor lowers the potential difference between the plates without a battery.
(2): The maximum electric field that a dielectric can withstand without causing it to break down is dielectric strength.

A
Both 1 and 2 are true, 2 is not correct explanation of 1

B
Both 1 and 2 are true, 2 is correct explanation of 1.

C
1 is false, 2 is true

D
1 is true, 2 is false

Solution

Consider a capacitor with charge density $$\sigma$$.
The potential between its two plates is given by $$\sigma d/ \epsilon_{0} $$
When a dielectric is inserted, the electric field inside the capacitor decrease decreasing the potential between the two plates of capacitor.
However, this is nothing to the dielectric strength of the dielectric.
Question 7
1 / -0

Two charges $$q$$ and $$-q$$ are kept apart. Then at any point on the perpendicular bisector of line joining the two charges:

A
the electric field strength is zero

B
the electric potential is zero

C
both electric potential and electric field strength are zero

D
both electric potential and electric field strength are non-zero

Solution

We know voltage is given by $$\dfrac{Kq}{r}$$
All the points on perpendicular bisectore are at equidistant from the two points A and B.
$$\therefore Potential = \dfrac{Kq}{r}+\dfrac{K(-q)}{r}$$
$$= 0$$
Question 8
1 / -0

A condenser is charged and then battery is removed. A dielectric plate is put between the plates of condenser, then correct statement is

A
Q is constant, V and U decrease

B
Q is constant, V increases U decreases

C
Q increases, V decreases U increases

D
Q, V and U increase

Solution

When a condenser is changed and the battery is removed the charge present on condenser remains same.
electric field is given by $$\dfrac{\sigma}{ {\epsilon}_0}$$
Now when dielectric plate is placed between $$C= \dfrac{K\epsilon_0A}{d}$$ , C increases
We know $$Q= C.V$$
$$V= \dfrac{Q}{C}$$
as C increases with addition of dielectric V decreases
$$\therefore $$ As V decreases U decreases
Question 9
1 / -0

When a dielectric material is introduced between the plates of a charged condenser, after disconnecting the battery, the electric field between the plates:

A
decreases

B
increases

C
does not change

D
may increase or decrease

Solution

We know electric field between capacitor plates: $$E= \dfrac{\sigma}{\epsilon _0}$$$$= \dfrac{V}{d}$$, where $$d$$ is the distance between the plates.

When battery is disconnected, the change present on capacitor is fixed, i.e., $$V$$ is constant.

Now, when the dielectric is introduced, the electric field becomes: $$E' = \dfrac{V}{Kd}$$, where K is dielectric constant which is always > 1

$$\therefore $$ electric field decreases.
Question 10
1 / -0

A dielectric of thickness $$5$$ cm and a dielectric constant $$10$$ is introduced between the plates of a parallel plate capacitor having plate area $$500 sq.$$ cm and separation between the plates $$10cm$$. The capacitance of the capacitor with the dielectric slab is $$\varepsilon _{0}=8.8\times 10^{-12}C^{2}/N-m^{2}$$

A
$$4.4 pF$$

B
$$6.2 pF$$

C
$$8 pF$$

D
$$10 pF$$

Solution

We know
$$C=\dfrac{kl_{0}A}{d}$$
effective capacitors is series combination of $$\dfrac{kl_{0}A}{5\ cm}\ and\ \dfrac{l_{0}A}{5\ cm}$$
$$\therefore \dfrac{1}{C_{eff}}=\dfrac{5\ cm}{kl_{0}A}+\dfrac{5\ cm}{kl_{0}A}$$

$$\Rightarrow C_{eff}=\dfrac{kl_{0}A}{(5+k5)\times 10^{-2}}$$

$$\Rightarrow C_{eff}=\dfrac{10\times 8.8\times 10^{-12}\times 500\times 10^{-4}}{55\times 10^{-2}}$$
$$\Rightarrow C_{eff}=8\ pF$$

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Electrostatic Potential and Capacitance Test - 28

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Electrostatic Potential and Capacitance Test - 28

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Question 1

A is true but B is false

B is true but A is false

Both A and B false

Both A and R are true

Solution

Question 2

potential becomes 1/k times

charge becomes k times

energy becomes 1/k times

electric intensity becomes k times

Solution

Question 3

$$10^{9}\ J$$

$$10^{6}\ J$$

$$9\times 10^{9}\ J$$

$$10^{10}\ J$$

Solution

Question 4

(a) and (b) are true

(a) and (c) are true

(b) and (c) are true

(b) and (d) are true

Solution

Question 5

$$90 V$$

$$9 V$$

$$900 V$$

$$270 V$$

Solution

Question 6

Both 1 and 2 are true, 2 is not correct explanation of 1

Both 1 and 2 are true, 2 is correct explanation of 1.

1 is false, 2 is true

1 is true, 2 is false

Solution

Question 7

the electric field strength is zero

the electric potential is zero

both electric potential and electric field strength are zero

both electric potential and electric field strength are non-zero

Solution

Question 8

Q is constant, V and U decrease

Q is constant, V increases U decreases

Q increases, V decreases U increases

Q, V and U increase

Solution

Question 9

decreases

increases

does not change

may increase or decrease

Solution

Question 10

$$4.4 pF$$

$$6.2 pF$$

$$8 pF$$

$$10 pF$$

Solution

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