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
420N/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 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})
R_{1} = 0 and R_{2}<(R_{4}-R_{3})
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)
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
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}
3 x 10^{4} C/m^{2}
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
3C_{1} = 5C_{2} = 0
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
50 sec and 100 sec
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
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}
-6aε_{0}
Let C be the capacitance of a capacitor discharging through a resistor R. Suppose t_{1} is the time taken for the energy stored in the capacitor to reduce to half its initial value and t_{2} is the time taken for the charge to reduce to one–fourth its initial value. Then the ratio t_{1}/t_{2} will be
1
1/2
1/4
1/4