Question

1. Give a comparison of the electrostatic and gravitational forces.

2. Four equal point charges each 16 micro coulomb are placed on the four corners of a square of side 0.2m. Calculate the force on any one of the charges.

3. Derive an expression for the electric field at any point on the Equatorial line of an electric dipole.

4. Why do the electrostatic field lines not form closed loops?

5. At point charge is placed at the centre of spherical Gaussian surface. How will electric flux change if

(1) the sphere is replaced by a cube of same or different volume

(2) a second charge is placed near, and outside, the original sphere,

(3) a second charge is placed inside the sphere and

(4) the original charge is replaced by an electric dipole?​

Answer 1

1. Electrostatic forces are much greater than gravitational forces . Gravitational forces are always attractive in nature, whereas electrostatic forces may be either attractive or repulsive. Gravitational forces are independent of the medium whereas electrostatic forces depend on the medium.
2. Let ABCD is square of side length 0.2m and four equal point charge each of 16 micro coulomb are placed on the four corner A, B, C and D of the square. we have to find force experienced by anyone of the charge . ... now, resultant of force exerted by B and D on A = √{(Kq²/a²)² + (Kq²/a²)²}= √2Kq²/a² , it is along CA .
3. The direction of Vector E1 is along PA. Now, let E2 be electric field intensity at P due to charge +q. ... Thus, the electric potential due to electric dipole is zero at every point on the equatorial line of the dipole. (b) The two identical charges are kept 2 m apart.
4. The electrostatic field lines do not form closed loop because no electric field line exist inside the charged body.
5. A point charge is placed at the centre of a closed Gaussian spherical surface of radius r. Electric flux passing through the surface is ψ. How is the electric flux ψ through the surface affected when the following changes are made in turn:(i) The spherical surface is replaced by a cylindrical surface of the same radius?(ii) The point charge is replaced by an electric dipole?Justify your answer is each case.

Answer 2

$\huge\colorbox{yellow}{Áηѕωєя࿐❤}$

1). єlєctrσstαtíc fσrcєs αrє much grєαtєr thαn grαvítαtíσnαl fσrcєs . grαvítαtíσnαl fσrcєs αrє αlwαчs αttrαctívє ín nαturє, whєrєαs єlєctrσstαtíc fσrcєs mαч вє єíthєr αttrαctívє σr rєpulsívє. grαvítαtíσnαl fσrcєs αrє índєpєndєnt σf thє mєdíum whєrєαs єlєctrσstαtíc fσrcєs dєpєnd σn thє mєdíum.

2). lєt αвcd ís squαrє σf sídє lєngth 0.2m αnd fσur єquαl pσínt chαrgє єαch σf 16 mícrσ cσulσmв αrє plαcєd σn thє fσur cσrnєr α, в, c αnd d σf thє squαrє. wє hαvє tσ fínd fσrcє єхpєríєncєd вч αnчσnє σf thє chαrgє . ... nσw, rєsultαnt σf fσrcє єхєrtєd вч в αnd d σn α = √{(kq²/α²)² + (kq²/α²)²}= √2kq²/α² , ít ís αlσng cα .

3). cσnsídєr αn єlєctríc dípσlє αв cσnsístíng chαrgєs +q αnd -q sєpαrαtєd вч dístαncє 2l αnd pσlє strєngth q. lєt, p вє thє pσínt σn thє єquαtσríαl línє αt dístαncє d frσm míd-pσínt σf thє dípσlє í.є. σ αs shσwn ín fígurє 1. nσw, thє єlєctríc fíєld, є 1

αt pσínt p duє tσ nσrth pσlє αlσng αp ís,

Rєfєr thє αttαchmєnt

4). thє єlєctrσstαtíc fíєld línєs dσ nσt fσrm clσsєd lσσp вєcαusє nσ єlєctríc fíєld línє єхíst ínsídє thє chαrgєd вσdч

5). frσm αвσvє,

(í) síncє thє chαrgє ínsídє thє gαussíαn surfαcє rєmαíns thє sαmє, thє єlєctríc fluх thrσugh ít rєmαíns unchαngєd.

(íí) síncє thє nєt chαrgє ínsídє thє surfαcє ís zєrσ, thє єlєctríc fluх pαssíng thrσugh thє surfαcє αlsσ вєcσmєs zєrσ.