Electric Potential Energy Potential And Dipoles(11th And 12th > Physics ) Questions and answers for exam Preparation

11th And 12th > Physics

ELECTRIC POTENTIAL ENERGY POTENTIAL AND DIPOLES MCQs

Total Questions : 30 | Page 1 of 3 pages

Three particles, each of charge $10 μ C$ are placed at the corners of an equilateral triangle of side 10 cm. The electrostatic potential energy of the system is
Given $(\frac{1}{4 π ϵ} = 9 \times 10^{9} N - m^{2} / C^{2})$

Zero
Infinite
27 J
100 J

Discuss Question

Answer: Option C. -> 27 J
:
C
For pair of charge

U = \frac{1}{4 π ϵ_{0}} [\frac{10 \times 10^{- 6} \times 10 \times 10^{- 6}}{\frac{10}{100}} + \frac{10 \times 10^{- 6} \times 10 \times 10^{- 6}}{\frac{10}{100}} + \frac{10 \times 10^{- 6} \times 10 \times 10^{- 6}}{\frac{10}{100}}]

= 3 \times 9 \times 10^{9} \times \frac{100 \times 10^{- 12} \times 100}{10} = 27 J

Question 2.

Four equal charges Q are placed at the four corners of a square of each side 'd'. Work done in removing a charge – Q from its centre to infinity is

0
$\frac{\sqrt{2} Q^{2}}{4 π ϵ_{0} a}$
$\frac{\sqrt{2} Q^{2}}{π ϵ_{0} a}$
$\frac{Q^{2}}{2 π ϵ_{0} a}$

Discuss Question

Answer: Option C. ->

\frac{\sqrt{2} Q^{2}}{π ϵ_{0} a}

:
C
Potential at centre O of the square

Four Equal Charges Q are Placed At The Four Corners Of A ...

V_{0} = 4 (\frac{Q}{4 π ϵ_{0} (a / \sqrt{2})})

Work done in shifting (– Q) charge from centre to infinity

W = - Q (V_{\infty} - V_{0}) = Q V_{0} = \frac{4 \sqrt{2} Q^{2}}{4 π ϵ_{0} a} = \frac{\sqrt{2} Q^{2}}{π ϵ_{0} a}

Question 3.

There is an electric field E along X-direction. If the work done on moving a charge 0.2 C through a distance of 2m along a line making an angle $60^{\circ}$ with the X-axis is 4.0, what is the value of E

$\sqrt{3} N / C$
4 N/C
5 N/C
None of these

Discuss Question

Answer: Option D. -> None of these
:
D

W = q V = q E . d

There Is An Electric Field E Along X-direction. If The Work ...

\Rightarrow 4 = 0.2 \times E \times (2 c o s 60^{\circ})

= 0.2 E \times (2 \times 0.5)

∴ E = \frac{4}{0.2} = 20 N C^{- 1}

Question 4.

Four identical charges $+ 50 μ C$ each are placed, one at each corner of a square of side 2m. How much external energy is required to bring another charge of $+ 50 μ C$ from infinity to the centre of the square $(G i v e n \frac{1}{4 π ϵ_{0}} = 9 \times 10^{9} \frac{N m^{2}}{C^{2}})$

64 J
41 J
16 J
10 J

Discuss Question

Answer: Option A. -> 64 J
:
A
Potential at the centre of square

V = 4 \times (\frac{9 \times 10^{9} \times 50 \times 10^{- 6}}{\sqrt{2}}) = 90 \sqrt{2} \times 10^{4} V

Work done in bringing a charge (q =

50 μ C

) from

\infty

to centre (O) of the square is

W = q (V_{0} - V_{\infty}) = q V_{0}

\Rightarrow W = 50 \times 10^{- 6} \times 90 \sqrt{2} \times 10^{4} = 64 J

Question 5.

Two insulated charged conducting spheres of radii 20 cm and 15 cm respectively and having an equal charge of 10 C are connected by a copper wire and then they are separated. Then

Both the spheres will have the same charge of 10 C
Surface charge density on the 20 cm sphere will be greater than that on the 15 cm sphere
Surface charge density on the 15 cm sphere will be greater than that on the 20 cm sphere
Surface charge density on the two spheres will be equal

Discuss Question

Answer: Option C. -> Surface charge density on the 15 cm sphere will be greater than that on the 20 cm sphere
:
C
After redistribution, charges on them will be different, but they will acquire common potential
i.e.

k \frac{Q_{1}}{r_{1}} = k \frac{Q_{2}}{r_{2}} \Rightarrow \frac{Q_{1}}{Q_{2}} = \frac{r_{1}}{r_{2}}

σ = \frac{Q}{4 π r^{2}} \Rightarrow \frac{σ_{1}}{σ_{2}} = \frac{Q_{1}}{Q_{2}} \times \frac{r_{2}^{2}}{r_{1}^{2}} \Rightarrow σ \propto \frac{1}{r}

i.e. surface charge density on smaller sphere will be more.

Question 6.

Two charges $+ 3.2 \times 10^{- 19}$ and $- 3.2 \times 10^{- 19}$ C placed $2.4 ˙ A$ apart form an electric dipole. It is placed in a uniform electric field of intensity $4 \times 10^{5}$ volt/m. The electric dipole moment is

$15.36 \times 10^{- 29} C o u l o m b \times m$
$15.36 \times 10^{- 19} C o u l o m b \times m$
$7.68 \times 10^{- 29} C o u l o m b \times m$
$7.68 \times 10^{- 19} C o u l o m b \times m$

Discuss Question

Answer: Option C. ->

7.68 \times 10^{- 29} C o u l o m b \times m

:
C
Dipole moment p = q (d)

= 3.2 \times 10^{- 19} \times (2.4 \times 10^{- 10}) = 7.68 \times 10^{- 29} C - m

Question 7.

An electron and a proton are at a distance of $1 ˙ A$ . The dipole moment will be $(C \times m)$

$1.6 \times 10^{19}$
$1.6 \times 10^{- 29}$
$3.2 \times 10^{19}$
$3.2 \times 10^{29}$

Discuss Question

Answer: Option B. ->

1.6 \times 10^{- 29}

:
B

p = q \times (d) = 1.6 \times 10^{- 19} \times 10^{- 10} = 1.6 \times 10^{- 29} C - m

Question 8.

Point charges q, q, -2q are placed at the corners of an equilateral triangle ABC of side l. The magnitude of net electric dipole moment of the system is

$q l$
$2 q l$
$\sqrt{3} q l$
$4 q l$

Discuss Question

Answer: Option C. ->

\sqrt{3} q l

:
C

Point Charges Q, Q, -2q Are Placed At The Corners Of An Equi...

P_{n e t} = \sqrt{p^{2} + p^{2} + 2 p p c o s 60^{\circ}} = \sqrt{3} p = \sqrt{3} q l

(∵ p = q l)

Question 9.

If an electron starts from rest from a point at which potential is 50 volt to another point at which potential is 70 volt, then its kinetic energy in the final state will be