The value of electric potential at a distance of 9 cm from the point charge 4 10-7 C is [Given 14 0 = 9 109 N m2 C-2]:
- 4 102 V
- 44.4 V
- 4.4 105 V
- 4 104 V
Answer
(4) 4 104 V
Solution
V = 14 0 qr
V = 9 109 4 10-79 10-2
V = 4 104 V
Answer: (4)
The value of electric potential at a distance of 9 cm from the point charge 4 10-7 C is [Given 14 0 = 9 109 N m2 C-2]:
(4) 4 104 V
V = 14 0 qr
V = 9 109 4 10-79 10-2
V = 4 104 V
Answer: (4)
The equivalent capacitance of the system shown in the following circuit is: [figure]
An electric dipole is placed as shown in the figure. [figure] The electric potential (in 102 V) at point P due to the dipole is ( 0 = permittivity of free space and 14 0 = K) :
The angle between the electric lines of force and the equipotential surface is
Two hollow conducting spheres of radii R1 and R2 (R1 >> R2) have equal charges. The potential would be
A capacitor of capacitance C = 900 pF is charged fully by 100 V battery B as shown in figure (a). Then it is disconnected from the battery and connected to another uncharged capacitor of capacitance C = 900 pF as shown in figure (b). The electrostatic energy stored by the system (b) is [figure]
The distance between the two plates of a parallel plate capacitor is doubled and the area of each plate is halved. If C is its initial capacitance, its final capacitance is equal to
The effective capacitances of two capacitors are 3 and 16 , when they are connected in series and parallel respectively. The capacitance of two capacitors are :
Six charges +q, -q, +q, -q, +q, and -q are fixed at the corners of a hexagon of side d as shown in the figure. The work done in bringing a charge q0 to the centre of the hexagon from infinity is ( 0 - permittivity of free space) [figure]
A parallel plate capacitor has a uniform electric field E in the space between the plates. If the distance between the plates is 'd' and the area of each plate is 'A', the energy stored in the capacitor is : ( 0 = permittivity of free space)
Polar molecules are the molecules :