Two point charges + 5 µC and − 2 µC a

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 Multiple Choice QuestionsMultiple Choice Questions

41.

Choose the wrong statement about equipotential surfaces.

  • It is a surface over which the potential is constant

  • The electric field is parallel to the equipotential surface

  • The electric field is perpendicular to the equipotential surface

  • The electric field is in the direction of steepest decrease of potential


42.

Three capacitors connected in series have an effective capacitance of 4 µF. If one of the capacitance is removed, the net capacitance of the capacitor increases to 6 µF. The removed capacitor has a capacitance of

  • 2 µF

  • 4 µF

  • 10 µF

  • 12 µF


43.

The electric potential V at any point (x, y, z) in space is given by V = 3x2 where x, y, z are all in metre. The electric field at the point (1 m, 0, 2 m) is

  • 6 V m-1  along negative x-axis 

  • 6 V m-1  along positive x-axis 

  • 12 V m-1  along negative x-axis 

  • 12 V m-1 along positive x-axis 


44.

Two equal point charges each of 3 µC are separated by a certain distance in metres. If they are located at (i + j + k)and (2i + 3j + k), then the electrostatic force between them is

  • 9 × 103 N

  • 9 × 10-3 N

  • 16.5 × 10-3 N

  • 3 × 10-3 N


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45.

Two resistors of resistances 200 kΩ and 1 MΩ respectively form a potential divider with outer junctions maintained at potentials of+ 3 V and −15 V. Then, the potential at the junction between the resistors is

  • + 1 V

  • − 0.6 V

  • Zero

  • − 12 V


46.

Three capacitors are connected in the arms of a triangle ABC as shown in figure. 5 V is applied between A and B. The voltage between B and C is

                        

  • 2 V

  • 1 V

  • 3 V

  • 1.5 V


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47.

Two point charges + 5 µC and − 2 µC are kept at a distance of 1 m in free space. The distance between the two zero potential points on the line joining the charges is

  • 27 m

  • 23 m

  • 2221 m

  • 2021 m


D.

2021 m

Let the potential be zero at P and Q.

Then solving for x1,

k × 5x1 = 2k1 - x1       x1 = 57

Similarly,

k × 51 + x2 = k × 2x2     x2 = 23

Separation PQ = 1 - 57 + 23

                      = 2021


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48.

C,V,U and Q are capacitance, potential difference, energy stored and charge of a parallel plate capacitor respectively. The quantities that increase when a dielectric slab is introduced between the plates without disconnecting the battery are

  • V and C

  • V and U

  • U and Q

  • V and Q


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49.

A capacitor of capacitance value 1 µF is charged to 30 V and the battery is then disconnected. If the remaining circuit is connected across a 2 µF capacitor, the energy lost by the system is

  • 300 µJ

  • 450 µJ

  • 225 µJ

  • 150 µJ


50.

Three charges Q0, -q and -q are placed at the vertices of an isosceles right angle triangle as in the figure. The net electrostatic potential energy is zero if Q0 is equal to

               

  • q4

  • 2q32

  • 2q

  • + q


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