# Physics Important Questions: Electrostatics

 Maharashtra State Board > Science >  Electrostatics > You are Here

### Very Short Answers (1 Mark)

Q1. Does the electric flux due to a point charge enclosed by a spherical Gaussian surface get affected when its radius is increased?

• According to Gauss’s law, flux through a closed surface is given by, Here, q is the charge enclosed by the Gaussian surface.
• Thus on increasing the radius of the gausGaussianface, charge q remains unchanged. So, flux through the Gaussian surface will not be affected when its radius is increased.

Q2. A charge Q  μC is placed at the centre of a cube. What would be the flux through one face?

• If a charge Q is placed in the centre of the cube, flux Φ through the complete surface of the cube is given by

Φ = Q/∈

• There are 6 faces of equal area for the cube. Thus the flux through each face is

Φ = Q/6∈

Where ∈ is the electrical permittivity of the medium.

Q3. An arbitrary surface encloses a dipole. What is the electric flux through this surface?

• An electric dipole consists of two equal oppositely charges separated by finite distance. Thus the total charge of the dipole is zero.
• Thus the arbitrary surface encloses a dipole means it encloses zero charge. Hence the electric flux through this surface is zero.

Q4. The capacitance of a charged capacitor is C and the energy stored in it is U. What is the value of charge on the capacitor?

The energy stored in the capacitor is given by

U = ½CV2  ………….. (1)

Where V is a potential difference across it

Now C = Q/V. Hence V = Q/C This is an expression for the value of charge on the capacitor.

Q5. Express dielectric constant in terms of capacitance with and without a medium. What is its unit?

• If Cm is the capacitance of a capacitor with a medium with dielectric constant ‘K’ between the plates and C is the capacitance of the same capacitor without medium between the plates. Then the dielectric constant is given by

K = Cm/C

• Dielectric constant is a pure ratio of similar quantities. Hence it is dimensionless, unitless quantity.

Q6. What is the function of a dielectric in a capacitor?

• The capacity of a parallel condenser is given by C = A∈0k/d.
• For particular condenser the area of the plate (A) and the distance between the plates (d) are constant. Similarly, electrical permittivity of free space (∈0) is constant.
• Thus the capacity of the condenser is directly proportional to the dielectric constant of the medium between the plates.
• Thus the capacity of the condenser can be increased by inserting a medium of a high value of dielectric constant.

Q7. How does the electric field inside a dielectric decrease when it is placed in an external electric field?

• The external field induces dipole moment by stretching or re-orienting the molecules of the dielectric. The net effect of these induced molecular dipole moment is the production of a field that opposes the external field.
• Thus dielectric gets polarised in opposite direction and, the electric field inside dielectric decreases.

Q8. How does the energy stored in a capacitor change (a) if the battery is disconnected b) if the plates of a charged capacitor are moved farther?

• if the battery is disconnected
• When only the battery is disconnected the charge on the plates and potential difference between the plates remains constant.
• Let us assume that after disconnecting the battery the charge on the plates is Q and the potential difference is V. Let C be the capacity of the capacitor, such that C = Q/V.
Thus energy stored in the capacitor before and after disconnecting battery is (1/2)×C×V2
• if the plates of a charged capacitor are moved farther
• The capacity of a parallel condenser is given by C = A∈0k/d. Thus when the distance between the plates increases, the capacity of the capacitor decreases.
• Thus the energy stored in the capacitor decreases.

Q9. How does the energy stored in a capacitor change, if the plates of a charged capacitor are moved farther, the battery remains connected?

• The capacity of a parallel condenser is given by C = A∈0k/d. Thus when the distance between the plates increases, the capacity of the capacitor decreases.
• The capacitor is still connected to the battery, means the potential difference between the plates is the same is not changing and remains constant (V).
• The energy stored in the capacitor is given by  (1/2)×C×V2. Thus the energy stored in the capacitor decreases.

Q10. When a capacitor is charged by a battery; is the energy stored in the capacitor same as energy supplied by the battery? why?

### S.A. I (2 Marks)

Q1.  Derive an expression for the electric field intensity at a point outside a charged conducting sphere.

• Consider a conducting sphere of radius R on which a charge +q is deposited. The deposited charge gets distributed uniformly over of the surface of the sphere. The lines of induction will be normal to the surface, radially outwards. Consider a point P at a distance r from the centre of the sphere (r > R) at which electric intensity is to be found. • To find the electric intensity at point P, construct an imaginary Gaussian spherical surface of radius ‘r’ through P.

The total Normal electric induction over element dS is given by

(TENI)dS  = k E εcos θ dS

As  electric intensity vector and area vector are parallel to eachother,

θ = 0° and hence cos θ  = cos 0° = 1

∴  (TENI)dS  = k E εo dS.

Total Normal Induction over whole Gaussian surface is This is the expression for an electric intensity at a point outside a charged sphere.

Q2. Derive an expression for the electric field intensity at a point outside an infinitely long charged cylindrical conductor.

• Consider a long uniform cylinder of radius ‘R’ carrying charge +q per unit length. Let ‘P’ be a point at a perpendicular distance ‘r’ from the axis (r > R) at which electric intensity is to be found. • To find the electric intensity at point P,  let us consider a coaxial imaginary Gaussian cylindrical surface of radius ‘r’ and height ‘l ‘ passing through the point P.
• If we consider an element dS on the flat face of the Gaussian cylinder, for this element the electric intensity vector and the area vector are perpendicular to each other. Hence θ = 90°, cos 90° = 0.

Hence TNEI over flat face = 0

• Therefore, the Total Normal Electric Induction over the whole surface is same as the TNEI over a curved surface. Consider an element dS over the curved surface of the Gaussian cylinder. Over this element,

(TENI)dS  = k E εcos θ dS

As  electric intensity vector and area vector are parallel to each other,

θ = 0° and hence cos θ  = cos 0° = 1

∴  (TENI)dS  = k E εo dS.

Total Normal Induction over whole Gaussian surface is As ‘q’ is the charge per unit length of the cylinder, the charge enclosed by the Gaussian cylinder is ‘ql’.

Applying Gauss’s theorem, This is the expression for an electric intensity at a point outside a charged cylinder.

Q3. Derive an expression for the electric field intensity at a point near a uniformly charged infinite plane sheet.

Q4. Assuming the equation for mechanical force per unit area of a charged conductor, obtain an expression for the energy density of a medium.

• During the process of charging a conductor, work has to be done to bring a charge on the surface of the conductor. This work is stored in the electric field surrounding the conductor in the form of electrostatic energy.A charged conductor is in an electric field.
• The mechanical force acting on it over dS area is given by The force is directed normally outwards. • Under the action of force, of the element dS moves outward through a distance dl, then the work done by the force is given by, • The work done dW is stored in the medium as electric potential energy dU. i.e. dW = dU. Then the energy per unit volume or energy density is given by This is an expression for energy per unit volume (energy density) of a medium.

Q5. Explain the concept of a capacitor.

Q6. Draw a neat labelled diagram of i) Cylindrical capacitor ii) Spherical capacitor.

Q7. Draw a neat labelled diagram of van de Graaff generator. Q8. Two isolated metal spheres A and B have radius R and 2R respectively, and same charge q. Find which of the two spheres has greater energy density just outside the surface of the spheres.
Q9. An infinitely long positively charged straight wire has a linear charge density
Cm-1 An electron is revolving around the wire as its centre with a constant velocity in a circular plane perpendicular to the wire. Deduce the expression for its kinetic energy.
Q10. What do you mean by polar molecules and non polar molecules?

• A polar molecule is one in which the centre of gravity of positive nuclei and revolving electrons do not coincide. Examples: HCl, H2O, N2O molecules.
• A non-polar molecule is one in which the centre of gravity of positive nuclei and revolving electrons coincide. Examples: O2, H2, CO2, Polyethene, polystyrene.

Q11. Derive an expression for electric field intensity at a point near and outside the surface of a charged conductor of any shape.

• Consider a uniform charged conductor of any shape. Let σ be the surface charge density or charge per unit area of its surface. Consider a point P very close to the surface of the conductor. • Consider an element dS near the point P. Construct an imaginary Gaussian surface of cross-sectional area dS with its axis normal to the surface, partly inside and partly outside the conductor. As the electric Intensity inside a conductor is zero, there is no TNEI through cross sectional area dS outside the conductor. Therefore TNEI through cross sectional area dS outside the conductor is

(TENI)dS  = k E εcos θ dS

As  electric intensity vector and area vector are parallel to each other,

θ = 0° and hence cos θ  = cos 0° = 1

∴  (TENI)dS  = k E εo dS.    ….  (1)

• Since is the charge density, the total charge enclosed by Gaussian surface over area dS is dS. Hence by Gauss’s theorem,

(TNEI) ds  = σ  dS    ….  (2)

From (1) and (2)

k E εo dS    =  σ  dS

∴   k E εo    =  σ

∴  E =  σ / k εo

This is the expression for an electric intensity at a point on near a charged conductor.

### S.A. – II (3 Marks)

Q1. State and prove Gauss’ theorem in electrostatics.

• Statement of Gauss’s Theorem : The Total Normal Electric Induction over any closed surface of any shape in an electric field is equal to the algebraic sum of electric charges enclosed by that surface.
• Explanation: Consider a closed surface enclosing number of charges such as +q1, + q2, – q3 …… then

(TNEI)S = q1 + q2 – q3 + ……= ∑ q.

#### Proof of Gauss’s Theorem : • Consider a charge +q situated at a point ‘O’ inside a closed surface of any shape. Such surface is called a Gaussian surface. Consider a small element dS on its surface. at a distance of ‘r’ from the charge ‘+q’.

The electric intensity  at any point over element dS is given by The normal drawn to the area dS makes an angle θ with the direction of electric intensity E.

Therefore, the total normal electric induction over area dS is given by The total Normal Electric Induction over the whole Gaussian surface

can be obtained by integrating above expression. The same argument is true for any other charge present inside the closed surface.

• Thus if the closed surface enclosed a number of charges such as +q1, + q2, – q3 …… then

(TNEI)S = q1 + q2 – q3 + ……= ∑ q

Thus Gauss’s theorem is proved.

Q2. Derive an expression for mechanical force per unit area of a charged conductor.

• Every element of a charged conductor experience a normal outward mechanical force. this is the result of repulsive force from similar charges present on the rest of the surface of the conductor.
• Consider a small element, dS on the surface of a charged conductor. If σ is the surface charge density and the charge carried by the element dS be dq.

dq =  σ .dS     ….(1) • Consider a point P just outside the surface near the element dS. The electric intensity at a point P is given by

E = σ /εok       ….(2)

The direction of the intensity is normally outwards. • Now, consider a point Q inside the conductor very near to element dS. • Here E2 is the electric intensity due to the charge on the rest of conductor. Hence repulsive force experienced by element dS carrying charge dq due to  is given by This is the mechanical force over dS area of a charged conductor. S.I.  Unit of force per unit area is  N/m².

Thus mechanical force per unit area is given by Q3. Derive an expression for the capacitance of a parallel plate capacitor completely filled with a dielectric. • A parallel plate condenser consists of two identical metallic plates P1 and P2. The plates are each of area A and distance ‘d’ apart. The space between two plates is filled with an insulating medium known as a dielectric. One of the plate (P1)is charged and the other plate (P2) is earthed.
• When a charge +Q is deposited on plate P1, an equal amount of negative charge, -Q is induced on earthed plate P2. This sets up a uniform electric field between the plates. The lines of force, therefore originate from P1 and terminate at P2.
• As the plate are very close to each other, the electric intensity near plates is given by,

E = σ /εok  ………. (1)

Where,   σ =  Charge per unit area  ( Surface charge density ), ε

εo  =  permittivity of free space.

k  =  dielectric constant of the medium.

• If ‘A’ the surface area of each phase, then the surface charge density is given by

σ =  Q /A   ……………… (2)

substituting for σ in equation (1) we get • The positive charge  +Q on plate P1 produce a potential difference V with respect to plate P2. Therefore, in the uniform electric field, the relation between intensity and potential is given by

E  =  V / d   ……………. (4)

From (3) and (4) we ge • This is an expression for the capacity of a parallel plate condenser with a dielectric. Therefore, the capacity of a parallel plate condenser increases with increase in surface area, dielectric constant but decrease in plate separation.

Q4. Derive an expression for the energy stored in a charged capacitor. Express it in different forms.

• A charged condenser stores in it an electrical potential energy (U) equal to the work (W) required to charge it.
• When a condenser acquires charge, its potential increases. The increases in potential is directly proportional to the amount of charge acquired.
• Consider a condenser of capacity C while charging, let v be its potential due to the deposition of charge ‘q’. The capacity of the condenser is given by • To deposit additional charge dq, work has to be performed against electric potential v. Hence work to be done (dW) to deposit an additional charge dq is given by The total work done to charge the condenser full i.e. from 0 to Q is given by This is an expression for energy stored in a charged condenser.

Q5. Derive an expression for the effective capacitance when three capacitors are connected in series.

• A number of condensers are said to be joined in series if they are joined end to end and each condenser acquires the same charge. In series combination, the plates of each condenser carry a charge of the same magnitude. However, the potential difference across the whole combination is the sum of the potential difference across individual condensers. • Consider three condensers of capacities C1, C2 and C3 respectively, then the total potential drop is given by :

V = V1 + V2 + V3   ….(1)

This is the terminal p.d. of the cell. In series combination, the charge on each condenser is same i.e. Q.

For the 1st condenser C1= Q/V1   Hence V1 = Q/C1

For the 2nd condenser, C2= Q/V2   Hence V2 = Q/C2

For the 3rd condenser, C3= Q/V3   Hence V3 = Q/C3

Substituting these values in equation (1)

V = Q/C1 + Q/C2 + Q/C3

∴  V = Q(1/C1 + 1/C2 + 1/C3)  …………….. (2)

• If the three condenses are replaced by an equivalent condenser of capacity ‘C’, then for equivalent or effective capacity of the condenser.

C = Q/V hence V = Q/C

Substituting in (2), we get,

Q/C = Q(1/C1 + 1/C2 + 1/C3)

∴  1/C = 1/C1 + 1/C2 + 1/C3

• In general, when ‘n’ condenser of capacity C1, C2, C3,…., Cn are connected in series, then their equivalent capacity is given by • Thus when a number of condensers are connected in series the reciprocal of their resultant capacity is the sum of reciprocals of their individual capacities.
• The resultant capacity of series combination is always less than the least capacity in the combination.

Q6. Obtain an expression for the equivalent capacitance when three capacitors are connected in parallel.

• A number of condensers are said to be joined in parallel if they are connected between two common points so that the potential difference across each condenser is same. However, each condenser acquires a charge depending on its capacity. • Let us consider three condensers of capacity C1, C2 and Cconnected in parallel across a cell of terminal potential difference  V. One plate of each condenser is connected to the positive terminal of the cell and the other plate is connected to the negative terminal. The total charge + Q supplied by the cell gets distributed over the condensers.
• If Q1, Q2 and Q3 are the amounts of charge accumulated by condenser C1, C2 and C3 respectively, then charge supplied by the cell is given by

Q   = Q1 + Q2 + Q3    ……….. (1)

In parallel combination p. d. across each condenser is same i.e. This is also the terminal p. d. of the cell.

For the 1st condenser C1= Q1/V  Hence Q1 = C1V

For the 2nd condenser, C2= Q2/V   Hence Q2 = C2V

For the 3rd condenser, C3= Q3/V   Hence Q3 = C3V

Substituting these values in equation (1)

Q = C1V + C2V + C3V

∴  Q = V(C1 + C2 + C3)

• If the three condensers are replaced by an equivalent condenser of capacity C, then for equivalent or effective condenser,

C = Q/V hence Q = CV

Substituting in (2), we get,

CV = V(C1 + C2 + C3)

C = C1 + C2 + C3

• In general, when ‘n’ condenser of capacities C1, C2, C3,  …… Cn are connected in series, then their equivalent capacity is given by • Thus when a number of condensers are joined in parallel, their effective capacity equals to the sum of the capacities of the individual condensers.
• The resultant capacity of parallel combination is always greater than the greatest capacity in the combination.

Q7. Explain polarisation of a dielectric in an external electric field.

• When polar or a non ploar dielectric are kept in the external electric field, their molecules acquire induced dipole moment and the dielectric is said to be polarised in the external electric field. The phenomenon is known as polarization.
• Polarization is defined as dipole moment per unit volume and is given by This relation is true for linear isotropic dielectrics.

• Linear isotropic dielectrics are those dielectrics in which induced dipole moment is induced in the same direction of the external field and is proportional to the field strength.

Q8. Describe the working and state the uses of the van de Graaff generator. • Working:
• The spray comb B1 is given a positive potential of range 104 V with respect to earth using high tension source. Due to the positive charge on the comb B1 electric wind is set and it sprays positive charge on the belt. As the belt moves and reaches the top of pulley P2. the negative charge is induced on sharp ends of collecting comb B2. The equal positive charge is induced on the far end of comb B2. Now, the far end of B2 and the sphere are connected. This positive charge on the far side of B2 moves immediately to the outer surface of the sphere.
• Due to discharging action of sharp edges of comb B2 negative electric winds are set and it discharges negative charge on the belt. Thus the positive charge on the belt is neutralized. This is repeated continuously. Thus positive charge gets accumulated on the outer surface of the sphere S.
• Now the capacity of the spherical conductor is given by Hence the potential of the sphere goes on increasing with the increase in the charge.

• Uses of Van De Graff Generator:
• It is used to create a high potential of a range of few million volts.
• It is used to accelerate projectiles like protons, deuterons etc.  which hit the target with large kinetic energy and bring about artificial transmutation.
• It is used to study collision experiments in physics
• Such beams are used for the treatment of cancer.
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