A combination of capacitors is set up as shown in the figure. The magnitude of the electric field, due to a point charge Q (having a charge equal to the sum of the charges on the 4 µF and 9 µF capacitors), at a point distant 30 m from it, would equal:
240N/C
360N/C
420N/C
480N/C
C.
420N/C
Resultant circuit,
As, charge on 3μF = 3μF x 8V = 24μC
Charge on 3μF = 3μF x 2V = 18 μC
charge on 4μF +Charge on 9μF
= (24 + 18)μC = 42μC
therefore,
Electric field at a point distant 30 m
A parallel plate capacitor is made of two circular plates separated by a distance of 5 mm and with a dielectric of dielectric constant 2.2 between them. When the electric field in the dielectric is 3 × 104 V/m, the charge density of the positive plate will be close to
6 x 10^{-7} C/m^{2}
3 x 10^{-7} C/m^{2}
3 x 10^{4} C/m^{2}
6x 10^{4} C/m^{2}
A.
6 x 10^{-7} C/m^{2}
When free space between parallel plate capacitor,
When dielectric is introduced between parallel plates of capacitor,
Electric field inside dielectric
where, K = dielectric constant of medium = -2.2
ε_{o} = permitivity of free space = 8.85 x 10^{-12}
σ = 2.2 x 8.85 x 10^{-12} x 3 x 10^{4}
= 6.6 x 8.85 x 10^{-8} = 5.841 x10^{-7}
= 6 x 10^{-7} C/m^{2}
The electrostatic potential inside a charged spherical ball is given by Φ = ar^{2} + b where r is the distance from the centre; a,b are constants. Then the charge density inside the ball is
-6aε_{0}r
-24πaε_{0}
-6aε_{0}
-24πaε_{0}r
C.
-6aε_{0}
Electric field,
A fully charged capacitor C with initial charge q_{0} is connected to a coil of self inductance L at t = 0. The time at which the energy is stored equally between the electric and the magnetic fields is
A.
As initially, charge is maximum
∴ q = q_{o}cos ωt
Current,
Assume that an electric field exists in space. Then the potential difference V_{A} – V_{O}, where VO is the potential at the origin and V_{A} the potential at x = 2 m is:
120 J
-120 J
-80 J
80 J
C.
-80 J
As we know, potential difference V_{A} -V_{o} is
dV = - Edx
A uniformly charged solid sphere of radius R has potential V0 (measured with respect to ∞) on its surface. For this sphere, the equipotential surfaces with potentials 3Vo/2, 5Vo/4, 3V/4 and Vo/4 have radius R_{1}, R_{2},R_{3} and R4 respectively. Then
R_{1} = 0 and R_{2}>(R_{4}-R_{3})
R_{1}≠0 and (R_{2}-R_{1})>(R_{4}-R_{3})
R_{1} = 0 and R_{2}<(R_{4}-R_{3})
2R<R_{4}
C.
R_{1} = 0 and R_{2}<(R_{4}-R_{3})
The potential at the surface of the charged sphere.
As potential decreases for outside points. Thus, according to the question, we can write
Similarly,
The figure shows an experimental plot for discharging of a capacitor in an R-C circuit. The time constant τ of this circuit lies between
150 sec and 200 sec
0 and 50 sec
50 sec and 100 sec
100 sec and 150 sec
D.
100 sec and 150 sec
Time constant τ is the duration when the value of potential drops by 63% of its initial maximum value (i.e, V_{o}/e)
Here, 37% of 25 V = 9.25 V which lies between 100s to 150 s in the graphs.
Charge +q and -q are placed at points A and B respectively which are distance 2L apart, C is the midpoint between A and B.The work done in moving a charge + Q along the semicircle CED is:
D.
work done is equal to change in potential energy.
In Ist case, when charge + Q is situated at C.
Electric potential energy of system
In IInd case, when charge + Q is moved from C to D.
Electric potential energy of system in that case
In the given circuit, charge Q2 on the 2 μF capacitor changes as C is varied from 1 μF to 3 μF. Q_{2} as a function of ‘C’ is given properly by : (figures are drawn schematically and are not to scale)
A.
From the given circuit,
Q_{c}= Q_{1}+Q_{2}
or
C (E-V) = 1 x V + 2 XV
Or V (C+3) = CE
As, C_{1} varied from 1μF to 3μF, charge increases with decreasing slope.
Two capacitors C_{1} and C_{2} are charged to 120 V and 200 V respectively. It is found that by connecting them together the potential on each one can be made zero. Then
5C_{1} = 3C_{2}
3C_{1} = 5C_{2}
3C_{1} = 5C_{2} = 0
9C_{1} = 4C_{2}
B.
3C_{1} = 5C_{2}
q_{1} = q_{2}
C_{1}V_{1} = C_{2}V_{2}
120C_{1} = 200C_{2}
⇒ 3C_{1} = 5C_{2}