If 2 charged particles having equal charges of 5.0 × 10-5C, each are brought from infinity to within a separation of 10 cm, then the increase in the electric P.E is
q1 = q2 = 5.0 × 10-5 C, r = 10 cm = 10-1 m
Electric P.E at infinite separation is zero.
Therefore, increase in P.E = P.E at this distance
= q1q2/4Πε0r = 9×10-9(5×10-5)(5×10-5)/10-1 = 225 J
A parallel plate capacitor has a capacitance of 5 μF. A slab of dielectric constant 2 is inserted between the plates and the capacitor is charged to 100 V and then isolated. If the dielectric slab is removed, the new P.D would be ____.
The capacitance of conductor on introducing dielectric slab,
C = 2 × 5 = 10 μF = 10 × 10-6
Therefore, q = CV = 10 × 10-6 × 100 (because V = 100 V) = 10-3 C
On removing the dielectric slab, the capacitance C1 = 5 μF.
Therefore, V1 = q/C1 = 10-3/5×10-6 = 200 V
If an electric dipole consists of 2 opposite charges, each of 1 μC separated by 3 cm, the dipole is placed in an external uniform field of 105 NC-1intensity, then the maximum torque exerted by the field on the dipole is
E = 105, q = 1 μC = 10-6 C, 2a = 3 cm = 0.03 m
Therefore, P = q (2a) = 10-6 × 0.03 = 3 × 10-8 cm
Now, τ = PE sinθ
Torque is maximum, when sin θ = 1
Therefore, τ max = 3 × 10-8 × 105 × 1= 0.003 Nm
A technician has only 2 capacitors. By using these singly, in series or parallel, he is able to obtain the capacitance of 3 μF, 4μF, 12 μF and 16 μF. What is the capacitance of the individual capacitors____.
Cp > C1
Therefore, Cp = 16 μF
Cs < C1 and Cs < C2
Therefore, Cs = 3 μF
Then, C1 = 4 μF + C2 = 12 μF
If electric dipole of charges 2 × 10-6 C, -2 × 10-6 C are separated by a distance 1 cm, then the electric field due to dipole at a point on its axial line 1 m from its centre is____.
q1 = 2 × 10-6 C, q2 = -2 × 10-6 C, 2d. = 1 cm = 10-2 m, r = 1 m
Electric field on the axial line,
E = 1/4Πε0•2p/r3 (because P = q × 2d)
because P = q × 2d and neglecting d in the denominator because d <<< r.
= 2 × 10-6 × 10-2 = 2 × 10-8
E = 9×109×2×2×10-8/13 = 360NC-1
An air capacitor of capacitance 3 μF is charged to 400V. If a medium with dielectric constant 3 is inserted between its plates and the potential is raised again to the same value, then the amount of energy stored will be ____.
C = 3 μF = 3 × 10-6 V = 400 V dielectric constant (k) = 3
U = 1/2 CV2 = 1/2 × 3 × 10-6 × 400 × 400 = 0.24 J (because)
On inserting the dielectric, C1 = KC = 3 × 3 × 10-6 = 9 × 10-6 F
Therefore, U1 = 1/2 C1V2 = 1/2×9×10-6×400×400 = 0.72 J
If the work done to move a charge of 0.2µC from one point to another point of 8 mJ, then the potential difference between the points is
V = w/q= 8×10-3/0.2×10-6= 4 x 104V
If a potential difference of 300 V is applied across the plates of a capacitor of 20 PF, then the charge on the plates will be
V = 300 V, C = 20 PF = 20 × 10-12 F
Q = CV = 20 × 10-12 × 300 = 6000 × 10-12 = 6 × 10-9 C
If 1200 J of work must be done to move an electric charge equal to 6 C from a potential of -20 V to V volts, then the value of V is
qo = 6 C, VA = -20 V, VB = V, WAB = 1200 J
VB - VA = WAB/qo
V - (-20) = 1200/6 V
V + 20 = 200 V
V = 180 V
Which of the following statement is correct regarding dielectric?
An insulator may not conduct electricity through it, but on applying electric field, induced charges are produced. Such an insulator is called dielectric. Here, the valence electrons in atoms of a dielectric are tightly bound to their nuclei.
The induced dipole moment of a non-polar molecule is
The alignment of the dipole moments of induced dipoles in the direction of applied electric field is called polarisation. Therefore, P ∝ E where P = induced dipole moment and E = external electric field.
Two condensers of capacity C1 and C2 are connected in parallel. If a charge q is given to the assembly, the charge gets shared. The ratio of the charge on the condenser C1 to the charge on the condenser C2 is ____.
Here, q1 = C1 V q2 = C2 V q1/q2 = C1/C2
The electrostatic potential is constant throughout the volume of the conductor because,
Electric field inside the conductor is zero. No work is done in moving a test charge on the surface of the conductor. Hence, there is no potential difference between any two points on the surface of the conductor. Therefore, the electrostatic potential is constant throughout the volume of the conductor.
How many farads gives 1 micro farad?
The s1 unit of C is farad.
If Q = 1 C and V = 1 V, then, C = 1 farad.
The smaller units of capacitance are 1 micro farad (1μF) = 10-6 farad, 1 nano farad (1nF) = 10-9 farad, 1 pico farad (1PF) = 10-12 farad.
What would be the potential at a point p due to a charge of 6 × 10-7located 9 cm away?
q = 6 × 10-7 C, r = 9 cm = 0.09 m
Potential at a point due to charge,
Vp = 1/4Πε0 ×q/r
= 9 × 109 × 6×10-7/0.09 = 6×104 V
What would be the magnitude of the charge if the electric potential at 0.2 m from a point charge is 60 V?
V = 1/4Πε0×q/r
q = 4πε0 ×Vr = 1/9×109 × 60 × 0.2
If electric dipole of charges 2 × 10-6 C, -2 ×106C are separated by a distance 1 cm, then the electric field due to dipole at a point on its equatorial line 1 m from its centre is
Electric field due to dipole at a point on the equatorial line,
E = P/4Πε0r3
But P = q × 2d = 2 × 10-6 × 10-2 = 2 × 10-8 cm
Therefore, E = 9×109×2×10-8/13 = 18×10-1=180NC-1
If 7 J of work is done in moving a positive charge of 0.7 C between two points, then potential difference between these two points is ____.
VB - VA = 7/0.7= 10 V
Surface charge density is greater than zero. The electric field then is____.
From the equation, E = (σ/ε0)n where σ = surface charge density, n = unit vector normal to the surface in the outward direction. For σ > 0, electric field is normal to the surface in outward direction.
What would be the energy stored in a condenser charged to a potential of 200 V with a charge of 0.1 C ____.
Energy U = 1/2 QV = 1/2 × 200 × 0.1 = 10 J
If a dipole is placed in a uniform electric field with its axis parallel to the field, then it experiences
Net force acting on the dipole
F = qE + (-qE) = 0
When θ = 0°
τ = pE sin0° = 0
Therefore, it experiences neither a net force nor a torque.
In a metal, the valence electrons ____.
In metallic conductors, the electrons are charge carriers. Thus in a metal, the valence electrons part away from their atoms and are free to move within the metal, but not free to leave the metal.
If an isolated spherical conductor has a capacity 2μF, then its radius would be ____.
C = 4Πε0r C = 2 μF = 2μF = 2 × 10-6 farad (1μF =10-6 farad)
r = C/4Πε0
Therefore, r = 2×10-6⧸1/9×109 = 2×10-6×9×109 = 18 × 103 m or 1.8 × 102 m
The maximum torque that acts on a molecule placed in an electric field 2.5 × 104 NC-1 having a dipole moment of 3.4 × 10-30 cm is
τ = PE sinθ
For maximum, θ = 90°
τ = 3.4 × 10-30 × 2.5 × 104 × sin 90°
= 3.4 × 10-26 × 2.5 × 4/4
= 34/4× 10-26 = 8.5 × 10-26 Nm
What happens to the excess charge when a conductor is charged?
By Gauss's theorem, there is no net charge enclosed by the closed surface. As volume and surface bounding it can be made as small as possible, there is no net charge at any point in the interior of the conductor. The excess charge, if any must reside on the outer surface of the conductor.
If 2 point charges and 20 μC and -20 μC are separated by a distance of 60 cm, then the electrostatic potential energy of the system will be ____. (Assume the zero of P.E to be at infinity)
U = (1/4Πε0)×(q1q2/r12 )
= (9×109)×(20×10-6×-20×10-6 )/0 . 6= -6 J
If 60 J of work must be done to move an electric charge equal to -200 C from A to B, then the potential difference will be
WAB = 60 J, q =-200 C
VB - VA= WAB/q (because WAB =q (VB - VA)
= 60/-200 = - 0.3V
The dielectric constant K of an insulator cannot be?
as E' = 0 for a conductor, K = ∞ for conductor. Thus, for an insulator K cannot be ∞.
If two charged particles having a charge of 2 × 10-5 C each, are brought from infinity to within a separation of 10 cm, then the increase in P.E during the process will be ____.
U=(1/4Πε0)(q1qr/r) = 9×109×2×10-5×2×10-5×100/10= 36 J
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