Maharashtra Board Class 11 Physics Chapter 10 Electrostatics Solution

Maharashtra Board Class 11 Physics Solution Chapter 10 – Electrostatics

Balbharati Maharashtra Board Class 11 Physics Solution Chapter 10: Electrostatics. Marathi or English Medium Students of Class 11 get here Electrostatics full Exercise Solution.


Maharashtra Class 11



Chapter Name


1.) Choose the correct option.

i.) A positively charged glass rod is brought close to a metallic rod isolated from ground. The charge on the side of the metallic rod away from the glass rod will be

(A) Same as that on the glass rod and equal in quantity

(B) Opposite to that on the glass of and equal in quantity

(C) Same as that on the glass rod but lesser in quantity

(D) Same as that on the glass rod but more in quantity


When two conductors are brought in contact with each other then charges are equally distributed of same nature among them.

ii.) An electron is placed between two parallel plates connected to a battery. If the battery is switched on, the electron will

(A) be attracted to the +ve plate

(B) be attracted to the -ve plate

(C) remain stationary

(D) will move parallel to the plates


If battery is switched on electric field is produced in between parallel plates having direction from positive plate to negative electron has negative charge so it will travel against the field that is towards the positive plate from negative plate.

iii.) A charge of + 7 µC is placed at the center of two concentric spheres with radius 2.0 cm and 4.0 cm respectively. The ratios of the flux through them will be

(A)1:4 (B) 1:2

(C)1:1 (D) 1:16


As charge is placed at center of two spheres so flux passing through first sphere will be same as passing through second sphere independent on size of sphere.

iv.) Two charges of 1.0 C each are placed one meter apart in free space. The force between them will be

(A) 1.0 N (B) 9×109 N

(C) 9×10-9 N (D) 10 N


v.) Two point charges of +5 µC are so placed that they experience a force of 8.0×10-3 N. They are then moved apart, so that the force is now 2.0×10-3 N. The distance between them is now

(A) 1/4 the previous distance

(B) double the previous distance

(C) four times the previous distance

(D) half the previous distance


According to coulombs law

F µ 1/r2

F1/F2= r22/r12

r22/r12= 8/2

r22/r12 = 4

r2/r1= 2

vi.) A metallic sphere A isolated from ground is charged to +50 µC. This sphere is brought in contact with other isolated metallic sphere B of half the radius of sphere A. The charge on the two spheres will be now in the ratio

(A) 1:2 (B) 2:1

(C) 4:1 (D) 1:1


When two conductors are brought in contact with each other than charges are equally distributed among them. Hence +25 µC is given to sphere B.

vii.) Which of the following produces uniform electric field?

(A) Point charge

(B) Linear charge

(C) Two parallel plates

(D) Charge distributed an circular any


Uniform electric field is produced by two parallel plates.

viii.) Two point charges of A = +5.0 µC and B = -5.0 µC are separated by 5.0 cm. A point charge C = 1.0 µC is placed at 3.0 cm away from the center on the perpendicular bisector of the line joining the two point charges. The charge at C will experience a force directed towards

(A) point A

(B) point B

(C) a direction parallel to line AB

(D) a direction along the perpendicular bisector


Given scenario is similar to a point situated on equatorial line of dipole so direction of force will be parallel to axis of dipole or line AB.

2.) Answer the following questions.

i.) What is the magnitude of charge on an electron?

Answer- 1.6 ´10-19c is the magnitude of charge on electron.

ii.) State the law of conservation of charge.

Answer-law of conservation of charge states that in any given physical process, charge may get transferred from one part of the system to another, but the total charge in the system remains constant” or, for an isolated system total charge cannot be created nor destroyed. In simple words, the total charge of an isolated system is always conserved.

iii.) Define a unit charge.

Answer- One coulomb is the amount of charge which, when placed at a distance of one metre from another charge of the same magnitude in vacuum, experiences a force of 9.0 ×109 N.

iv) Two parallel plates have a potential difference of 10V between them. If the plates are 0.5 mm apart, what will be the strength of electric charge.


Strength is given by

E = V/d

E = 10/(0.5  x 10-3)


E = 10/0.5 x 103

E = 20 x  103 V/m.

v.) What is uniform electric field?

Answer- the electric field in which magnitude as well as direction of field remains same at all the points in the field is called uniform electric field.

vi.) If two lines of force intersect of one point. What does it mean?

Answer- line of forces represents the direction of electric field so two line of forces at a single point represents two different directions of electric field at that point.

vii.) State the units of linear charge density.

Answer-linear charge density is charge per unit length so has unit of coulomb per meter or (c/m).

viii.) What is the unit of dipole moment?

Answer-the dipole moment is given byP= 2qlHence has unit of coulomb-meter or c-m,

ix.) What is relative permittivity?


Relative permittivityεr is the ratio of absolute permittivity of a medium to the permittivity of free space.

εr=  ε/ε0 .


Relative permittivity (εr) is the ratio of the force between two point charges placed a certain distance apart in vacuum to the force between the same two point charges when placed at the same distance in the given medium.

εr = Fvacuum/Fmedium .

Updated: April 23, 2022 — 11:46 pm

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