Q: What is meant by the statement "Charge is quantized"? (2 Marks) A: The electric charge on any charged body can be expressed as an integral multiple of charge of electron i.e., Q = ± ne where 'n' is an integer, e = 1.6 × 10−19 C. This is known as Quantization of electric charge.
Physics G.V. Chandra Sekhar
Electric Charges and Fields 4 Marks Questions Q: State and explain Coulomb's inverse square law in electricity. A: Coulomb's inverse square law: The force of attraction or repulsion between two stationary point electric charges is directly proportional to the product of the magnitudes of two charges and is inversely proportional to the square of the distance between them. The force acts along the straightline joining the two charges. ★ Explanation: Let us consider two stationary point electric charges Q1 and Q2 separated by a distance 'r' from each other with air or vacuum as the medium between them. Let 'F' be the mutual force of attraction or repulsion between them. F ∝ Q1Q2 1
F
Q1 →
Q1Q2 ∴ F ∝ ⎯⎯ r2 1 Q1Q2 (OR) F = ⎯⎯ . ⎯⎯ 4πε0 r2 where ε0 = permittivity of free space. = 8.85 × 10−12 C2/N.m2 1 ⎯⎯ = 9 × 109 N.m2/C2 4πε0 Fore between the two charges depends on nature of the medium between them. If the space between the charges is completely filled by a material, then force between them is, Q1Q2 1 Fm = ⎯⎯ . ⎯⎯ 4πε r2 where ε = permittivity of material = ε0 εr, εr is relative permittivity of material. Q: Derive an expression for the moment of couple acting on an electric dipole in a uniform electric field. A: Consider an electric dipole consisting of two electric charges '+ Q' and '−Q' separated by a distance '2l'. Let the dipolemoment of the
2 Marks Questions Q: Repulsion is the sure test of electrification than attraction. Why? A: A charged body may attract a neutral body or an opposite charged body. But it always repels a like charged body. Hence Repulsion is the sure test of electrification. Q: How many electrons constitute in 1C of electric charge? A: Electric charge, Q = ± ne Q 1C ∴ n = ⎯ = ⎯⎯⎯⎯⎯⎯ e 1.6 × 10−19 C = 6.25 × 1018 electrons. Q: The electric lines of force do not intersect. Why? A: If they intersect at the point of intersection intensity of electric field will have two different directions, which is not possible. So they do not intersect. Q: What happens to force between two electric charges, if the distance between them is (a) halved & (b) doubled. 1 A: From Coulomb's inverse square law F ∝ ⎯ d2 F1 d2 2 F1 d1/2 2 1 (a) ⎯ = ⎯ ⇒ ⎯ = ⎯⎯ = ⎯ F2 d1 F2 d1 4
( )
dipole be P = Q × 2l. Let the dipole be placed in a uniform electric field of intensity 'E' such that the angle made by the axis of dipole with the direction of electric field is 'θ' as shown. QE →
( )
⇒ F2 = 4F1 ∴ The force between the charges becomes
+Q Β
θ
θ
0 −Q A
N
d
QE ★ Force on '+Q' charge is 'QE' in the direction → of E. Force on '−Q' charge is 'QE' in opposite → to the direction of E. These two equal and opposite parallel forces form a couple. This couple tries to rotate the dipole parallel to the electric field. Moment of couple, τ =
(Force on either of electric charges)
(Perpendicular distance between the forces)
×
⇒ τ = QE × d
AN d Form ∆ANB, sinθ = ⎯ = ⎯ AB 2l (... AB = 2l = length of dipole) ⇒ d = 2l × sinθ ∴ τ = QE × 2lsinθ
= (Q × 2l) . E sinθ .. = P × E × sinθ ( . P = Q × 2l) → → In the vector form, → τ = P × E Q: Derive an expression for the intensity of electric field at a point on the axial line of an electric dipole. A: Consider an electric dipole consisting of two '4' times the initial value. F1 (b) ⎯ = F2
d2 2 F1 2d1 2 F1 ⎯ ⇒ ⎯ = ⎯⎯ ⇒ F2 = ⎯ d1 F2 d1 4
( )
( )
∴ Force between the charges is decreased to 1/4th of initial value. Q: State Gauss's law in electrostatics. A: The total electric flux through any closed 1 surface is equal to ⎯ times the total electric
ε0
charge enclosed by the closed surface.
→ → QTot ∴ Electric flux, φ = Ο∫ E. d s = ⎯⎯ s
electric charges +Q and −Q separated by a distance 2l. Let 'A' be a point on the axial line of dipole at a distance 'r' from its centre.
. O
−Q
E
← Q2
⏐←⎯ r ⎯→⏐
and F ∝ ⎯ r2
Gauss's law in electrostatics?
ε0
Q: Write an expression for electric intensity due to infinite long charged wire, at a radial distance 'r' from the wire. λ A: E = ⎯⎯ 2πε0r where λ = linear charge density = electric charge per unit length. ε0 = permittivity of free space. Q: Write the expression for electric intensity due to an infinite plane sheet of charge. σ A: E = ⎯⎯ 2ε0 where σ = surface charge density = electric charge per unit surface area.
ε0 = permittivity of free space.
→ Ε
→ P
>
(−)
<
+Q
.
A
⎯→ l ←⎯⎯→ l ←⎯ ⎯⎯⎯⎯⎯⎯⎯→ r ←⎯⎯⎯⎯ Electric field intensity at point 'A', 1 Q due to '−Q' charge is, E(−) = ⎯⎯ . ⎯⎯⎯ 4πε0 (r + l)2
→ (along AO) 1 Q due to '+Q' charge is, E(+)= ⎯⎯ . ⎯⎯⎯ 4πε0 (r − l)2 → (along OA) ∴ Resultant electric field intensity at point 'A" is, → Ea = E(+) − E(−) (along OA) 1 Q 1 Q ⇒ Ea= ⎯⎯ . ⎯⎯⎯ − ⎯⎯ . ⎯⎯ 4πε0 (r − l)2 4πε0 (r + l)2 Q = ⎯⎯ 4πε0 Q = ⎯⎯ 4πε0 Q = ⎯⎯ 4πε0
[ [
1 1 ⎯⎯⎯ − ⎯⎯ (r − l)2 (r + l)2
(r + l)2 − (r − l)2 ⎯⎯⎯⎯⎯ (r − l)2 × (r + l)2
[
4rl ⎯⎯ ⎯ 2 (r − l2)2
]
]
]
1 2.p.r ∴ Ea = ⎯⎯ . ⎯⎯⎯⎯ 4πε0 (r2 − l2)2 (along the axial line of dipole from '−Q' to '+Q' charge) ∼ r2 For a short dipole, l << r so r2 − l2 − 1 2.p.r ∴ Ea= ⎯⎯ . ⎯⎯ 4πε0 (r2)2
. θθ
→ Ε
→ Ε
(+)
e
→ r +Q B
Ε (−)
θ → P 0
l
−Q
l
C
Electric field intensity at point 'A', Due to '+Q' charge is, 1 Q E(+) = ⎯⎯ . ⎯⎯ 4πε0 (BA)2 1 Q = ⎯⎯ . ⎯⎯⎯ ⎯⎯ 4πε0 2 r2 + l2 .. ( . (OB)2 + (OA)2 = (BA)2) → 1 Q ⇒ E(+) = ⎯⎯ . ⎯⎯ (along BA) ......(1) 4πε0 r2 + l2 Due to '−Q' charge is, 1 Q E(−) = ⎯⎯ . ⎯⎯ 4πε0 (CA)2 1 Q = ⎯⎯ . ⎯⎯⎯⎯ ⎯⎯⎯ 2 4πε0 2 r + l2 (... (OC)2 + (OA)2 = (CA)2)
)
(√
1 Q ⇒ E(−) = ⎯⎯ . ⎯⎯ 4πε0 (r2 + l2)
)
→ (along AC) ......(2)
From (1) & (2) E(+) = E(−) (only in magnitude) 1 Q Let E(+) = E(−) = E = ⎯⎯ . ⎯⎯⎯ 4πε0 (r2 + l2)
1 2.p.r = ⎯⎯ . ⎯ ⎯ 4πε0 r4
∴ Resultant electric field intensity at point 'A' is, 2θ Ee = 2.E.cos ⎯ = 2.E.cosθ 2
( )
2p
⇒ Ea = ⎯⎯ . ⎯ r3
l
Q: State Gouss's law in electrostatics and its importance. A: Gouss's law: The total electric flux through 1 any closed surface is equal to ⎯ times the
ε0
net electric charge enclosed by the closed surface. Q ∴ Electric flux φE = ⎯
ε0
→ → Q ⇒ Ο∫ E . d s = ⎯ s
A
(√
1 2 × (Q × 2l) × r = ⎯⎯ . ⎯⎯ ⎯ 4πε0 (r2 − l2)2 But dipolemoment of the dipole, P = Q × 2l (along the axis of dipole from −Q to +Q)
1 4πε0
(ii) Symmetric considerations in many problems become very easy by the application of Gauss's law. Q: Derive an expression for the intensity of the electric field at a → → Ε(+) Εa point on the equatorial line of an >> > electric dipole. A: Consider an electric dipole consisting of two electric charges '+Q' and '−Q' separated by a distance '2l'. Let 'A' be a point on its equatorial line at a distance 'r' from its centre.
ε0
Where Q = Total charge enclosed by the closed surface ε0 = permittivity of free space Importance: (i) By using Coulomb's law and the principle of superposition we can find electric field strength at a given point. But in the case of more complex configurations to charge, the electric field intensity can more easily calculated by applying Gauss's law.
l
But in ∆AOB, cosθ = ⎯ = ⎯⎯⎯⎯ ⎯⎯⎯ AB ( r2 + l2)
√
1 Q l ∴ Ee = 2 . ⎯⎯ . ⎯⎯ . ⎯⎯⎯ ⎯⎯ 4πε0 (r2 + l2) r2 + l2 1 Q × 2l = ⎯⎯ . ⎯⎯⎯ 4πε0 (r2 + l2)3/2
√
(Parallel to axis from +Q to −Q) For a short dipole, l << r so (r2 + l2) ≈ r2 1 P ∴ Ee = ⎯⎯ . ⎯ (... P = Q × 2l) 4πε0 r3
For Intermediate study material, model papers visit... www.eenadupratibha.net Mail your feedback to..
[email protected]
