### Electrostatics Objectives Part 01

01 . Permittivity is expressed in farad/m.

02 . Absolute permittivity of vacuum is taken as 8.854*10^-12

03 . Relative permittivity of vacuum is unity.

04 . A charge which when placed in vacuum from an equal and similar charge repels with a force of 9000N, is known as micro-coulomb.

05 . Force between two charged particles is proportional to inverse square of distance.

06 . Electric field intensity is a vector quantity.

07 . Unit of electrostatic flux density is coulomb per square meter.

08 . By increasing the area of overlap between the plates of a capacitor when keeping voltage across plates const, energy increases.

09 . A force of 1N is experienced between two equal charges in space, separated by 1 meter and having a magnitude of 10 micro coulombs.

10 . A point charge in space is attracted towards a dielectric material because of the max electrostatic flux.

11 . Point charges 30nC, -20Nc and 10Nc are located at (-1, 0, 2), (0, 0, 0) and (1, 5, -1) respectively. The total flux leaving a cube of side 6 meter centered at the origin is 10nC.

12 . A positive and a negative charge are initially 4 cm apart. When they are moved closer together so that they are now only 1 cm apart, the force between them will be 16 times larger than before.

13 . Air has least dielectric strength.

14 . The force of attraction or repulsion between two charges q1 and q2 at a distance d meters apart is proportional to the product of charges and inversely proportional to the distance square between the two charges.” Above statement is attributed to Coulomb’s Law.

15 . The unit of intensity is N/C.

16 . The electric field on a plane is described by V=20[1/r + 1/r*r]. The field is due to a dipole and a monopole.

17 . Two capacitors of each breakdown voltage 500V are connected parallel. The breakdown rating of the combination will be 500V.

18 . A plane Z=10m carries charge 20nc/m^2. The electric field intensity at the origin is –360piAx/V/m.

19 . Two capacitors each of breakdown voltage 250V are connected in series. The breakdown voltage of the combination will be 500V.

20 . The work done by force F = 4Ax – 3Ay + 2Az. Newton in giving a 1nc charge a displacement of D = 10Ax + 2Ay – 7Az is 20nJ.