Electrostatic Potential And Capacitance

Given,

Two charges q_A = 5 x 10^-8 C and q_B = -3x10^-8 C

Let O be the center of the hexagon. It contains the charges at all its 6 vertices, each charge = + 5 μC = 5 × 10^-6 C. The side of the hexagon is 10 cm = 0.1 m

It follows that the point O, when joined to the two ends of a side of the hexagon forms an equilateral triangle

Electric potential at O due to one charge,

Here, q = amount of charge = 5 × 10^-6 C

r = distance between charge and O = 0.1 m

ε₀ = absolute permittivity of free space

Since at the each of the hexagon, a charge of 5 μC (5 × 10^-6 C) is placed, total electric potential at the point O due to the charges at the six corners,

⇒

⇒

⇒ V₁ = 2.7 × 10⁶ V

(a) For the given system of two charges, the equipotential surface will be a plane normal to the line AB joining the two charges and passing through its mid-point O. On any point on this plane, the potential is zero.

(b) The electric field is in a direction from the point A to point B i.e. from the positive charge to the negative charge and normal to the equipotential surface.

Given, q = 1.6 x 10^-7 C

Radius of the sphere, r = 12 cm = 0.12 m

(a) Inside the sphere: The charge on a conductor resides on its outer surface. Therefore, electric field inside the sphere is zero.

(b) Just outside the sphere: For a point on the charged spherical conductor or outside it, the charge may be assumed to be concentrated at its center.

Therefore, the electric field just outside the sphere is given as:

where, q is the charge due to which the field is being calculated

putting the values of “r” and “q” in the equation, we get,

∴E = 10⁵ N/C

(c) At a point 18 cm from the center: Given, R = 18 cm

Electric field at a point 18 cm from the center

∴ E = 4.4 x 10⁴ N/C

Given: Capacitance of capacitor when medium between two plates is air, C = 8 pF = 8 × 10^–12 F

Now, the capacitance of a parallel plate capacitor is given as:

where, A be the area of each plate

d be the distance between the two plates of the parallel plate capacitor

k is the dielectric constant, (for air , k = 1)

Capacitance, C = 8 pF(Given)

⇒

Suppose that the capacitance of the capacitor becomes C’ when the distance between the plates is reduced to half (d’ = d/2) and the space between them is filled with a substance of dielectric constant K = 6.

⇒

From equation (i) and (ii),

C’ = 12 x 8 x 10^-12 = 96 x 10^-12 = 96 pF

Given, C₁ = C₂ = C₃ = 9 pF = 9 x 10^-12 F

V = 120 volt

(a) Total capacitance of the series combination is given by

⇒

⇒

C_net = 3 x 10^-12 F = 3 pico-Farad.

(b) Let q be the charge on each capacitor.

Sum of the potential difference across the plates of the three capacitors must be equal to 120 V

⇒ V₁ + V₂ + V₃ = V_net

⇒ (∵ V_net = 120 V)

⇒

⇒

⇒ q = 360 x 10^-12 C

Therefore, potential difference across a capacitor

Given, C₁ = 2 pF

C₂ = 3 pF

C₃ = 4 pF , V = 100 volt

(a) Total capacitance of the parallel combination is

C = C₁ + C₂ + C₃ = 2 + 3 + 4 = 9 pF

(b) Let q₁ , q₂ and q₃ be that charges on the capacitor C_{1 ,} C₂ and C₃ respectively.

In the parallel combination the potential difference across each capacitor will be equal to the supply voltage i.e., 100 V

⇒ q₁ = C₁V = 2 x 10^-12 × 100 = 2 × 10^-10 C

⇒ q₂ = C₂V = 3 x 10^-12 × 100 = 3 × 10^-10 C

⇒ q₃ = C₃V = 4 x 10^-12 × 100 = 4 × 10^-10 C

Given, A = 6 x 10^-3 m²

D = 3 mm = 3 x 10^-3 m

Capacitance of the capacitor,

⇒ C = 1.77 x 10^-11 F = 17.7 pF

When the capacitor is connected to a 100 V supply, the charge on the each plate of the capacitor,

q = CV = 1.77 x 10^-11 x 100 = 1.77 x 10^-9 C

(a) When the voltage supply remains connected:

The capacitance of the capacitor will become K times.

Therefore, C’ = kC

Where k = dielectric constant = 6 × 17.7pF = 106.2 pF

The potential difference across the two plates of the capacitor will remain equal to the supply voltage i.e. 100 V

The charge on the capacitor,

q’ = C’V = 160.2 x 10^-12 x 100

= 1.602 x 10^-8 C

(b) After the voltage supply is disconnected:

As calculated above, the capacitance of the capacitor, C’ = 106.2 pF

The potential difference will decrease on introducing mica sheet by a factor of K,

⇒

The charge on the capacitor,

q’ = C’V’ = 106.2 × 10^-12 × 16.67

q’ = 1.77 x 10^-9 C

Given, C = 12 pF = 12 x 10^-12 F

V = 50 V

The electrostatic energy stored in the capacitor,

W = (1/2) CV² = (1/2) × 12 × 10^-12 × (50)² = 1.5 × 10^-8 J

Given, C₁ = 600 pF = 600 x 10^-12 F

V₁ = 200 V

Energy stored in the capacitor,

U₁ = (1/2) C₁ (V₁)² = (1/2) × 600 × 10^-12 × (200)²

= 12 × 10^-6 J

When this charged capacitor is connected to another uncharged capacitor C₂ ( = 600 pF) ,they will share charges, till potential differences across their plates become equal.

Total charge on the two capacitors,

q = C₁V₁ + C₂V₂ = 600 × 10^-12 × 200 + 0

= 12 × 10^-8 C

Total capacitance of the two capacitors,

C = C₁ + C₂ = 600 pF + 600 pF

= 1200 pF

= 1200 x 10^-12 F

Therefore, common potential of the two capacitors,

Energy stored in the combination of the two capacitors,

U₂ = (1/2)CV² = (1/2) x 1200 x 10^-12 x (100)²

= 6 x 10^-6 J

Therefore, energy lost by the capacitor C₁( in the form of heat and electromagnetic radiation),

U₁ – U₂ = 12 x 10^-6 – 6 x 10^-6 = 6 x 10^-6 J

Given,

Charge at origin (O), q₀ = 8 × 10^-3 C

The charge (q₁ = -2 Nc) is moving through the given points, P(0, 0, 3)cm, R(0, 6, 9)cm and Q(0, 4, 0)cm respectively.

The picture below represents above situation,

Since, Work is independent on the path followed we concentrate on point P and Q only.

Potential at any point is given by,

Where,

q = charge

d = distance from origin

ϵ₀ = permittivity of space

and,

Thus, Potential at point P,

(∵ distance of point from origin = 3 cm = 0.03m)

Potential at point Q,

Work done, W = q₁ × (Change in potential difference)

= 1.27 J

Given, side of the cube = b units

Charge at each vertex = q C

In the cube shown above,

d = diagonal of any face = = b units

l = diagonal of whole cube = = √3 b units

Thus, the distance of centre of the cube from each vertices = = units

∴ Electric potential at the centre of the cube, V =

(∵ 8 charges)

Where,

q = charge

d = distance from origin=

ϵ₀ = permittivity of space

∴ V = N units C^-1

As the charges as symmetrically placed along the vertex of the cube, the net Electric field will be zero at the centre as the field due to 2 charges placed opposite to each other will cancel, so for all charges, net field will be 0.

Potential at any point is given by, V = N m C^-1

Electric field generated by a charge, E = N C^-1

Where,

q = charge

d = distance from origin

ϵ₀ = permittivity of space

and, = 9 × 10⁹ N m² C^-2

(a) Figure below represents the given situation,

Let the potential at O (midpoint) be V and Electric field be E.

We have,

∴ V = + = 2.4 × 10⁵ N m C^-1

and,

E = electric field due to charge 2.5μC – electric field due to charge 1.5μC

∴E =

⇒ E = 4 × 10⁵ N C^-1

(b) Figure below represents the given situation,

From the figure point Z is required location and its distance from both the charges is equal and equal to,

d = AZ = BZ =

⇒ d = 0.18 m

and θ =

∴ θ = 56.25⁰

Thus, Potential and electric field at point Z is given by,

V =

⇒ V = 2 × 10⁵ N m C^-1

and,

E =

Where,

E₁ = electric field generated by 1.5μC charge at Z

E₂ = electric field generated by 2.5μC charge at Z

and, θ = angle between distance and perpendicular height (as shown above) and,

∴ E = 6.6 × 10⁵ N C^-1

(a) Charge placed at the centre of the sphere is + q. Thus, charge induced on the inner surface be –q.

Surface charge density at the inner surface of the sphere,

σ₁ = =

Charge placed at the centre of the sphere is + q. Thus, charge induced on the inner surface be + q and surface charge distribution is Q. Therefore Total charge = Q + q.

Surface charge density at the outer surface of the sphere,

σ₂ = =

(b) Yes

Whatever the shape of the shell, the electric field intensity in the cavity is zero. Take a closed loop such that it is inside the cavity and in conductor. Net work done in carrying charge along closed loop by electric field is zero. Thus, electric field is zero, whatever the shape.

Electric field on one side of the charged is body is E₁ and the electric field on the other side of the same body be E₂. If infinite plane charged body has uniform thickness then the electric field due to one surface of the body is given by,

=

Where,

= a unit vector normal to the surface at a point

σ = surface charge density at that point.

ϵ₀ = permittivity of free space.

Electric field due to other surface of the charged body,

=

∴ Electric field due to both surfaces,

- = + =

Since, Inside a closed conductor electric field s zero,

∴ = =

Hence, the electric field just outside the conductor =

Net work done in carrying charge along closed loop by electric field is zero. Hence, the tangential component of the electric field is continuous from one side of a charged surface to the other.

Given,

Charge density of the long-charged cylinder of length L and radius r is λ. Another same type of cylinder with radius R surrounded it.

Let E is the electric field produced in the space between the two cylinders.

Electric flux through a Gaussian surface is given by the Gaussian theorem as,

Φ = E(2πd)L

Where, d = distance of a point from common axis of the cylinders.

Let q be the total charge on the cylinder,

∴ Φ = E(2πd)L =

Where, q = charge of the inner sphere of the outer cylinder

ϵ₀ = permittivity of space

Thus,

E(2πd)L =

∴

Hence, the electric field between cylinders,

Given,

The distance between electron-proton of hydrogen atom = 0.53 Å = 0.53 × 10^-10m

Charge on electron, q₁ = - 1.6 × 10^-19 C

Charge on proton, q₂ = 1.6 × 10^-19 C

(a) Potential energy at infinity is Zero.

Potential energy of a system,

= Potential energy at infinity – Potential energy at distance d

= 0 -

Where, ϵ₀ = permittivity of space

and, =

∴ Potential energy,

⇒ V = -43.7 × 10^-19 J

On converting into electron volts we get,

⇒

⇒ V = -27.2 eV

(∵ 1eV = 1.6 × 10^-19 J)

(b) Given, Kinetic energy is half of the potential energy in (a).

∴ Kinetic energy, K = 0.5 × (27.2) = 13.6 eV

Thus, Total energy of electron = -27.2 + 13.6 = -13.6 eV

Amount of work require to free the electron, A = Increase in energy the electron.

∴ A = 0-(-13.6) = 13.6 eV

(c) If we take potential energy Zero at 1.06 Å separation, then the potential energy of the system,

P =

⇒P = -21.74 × 10^-19 J

In electron volts, the potential energy is given as,

P = -13.585 eV.

(1eV = 1.6 × 10^-19 J)

∴ The amount of work done to free the electron in this case,

W = -27.17- (-13.585) = -13.585 eV

The system of two protons and electron is represented as below,

The total potential energy at infinity is Zero.

Thus, Potential energy of the system is,

V =

Where,

Charge on proton 1, q₁ = 1.6 × 10^-19 C

Charge on proton 2, q₂ = 1.6 × 10^-19 C

Charge on electron, q₃ = - 1.6 × 10^-19 C

Distance between proton 1 and proton 2, d₁ = 1.5 × 10^-10 m

Distance between proton 1 and electron, d₂ = 1 × 10^-10 m

Distance between proton 2 and electron, d₃ = 1 × 10^-10 m

ϵ₀ = Permittivity of free space

and,

∴ V =

⇒ V = -30.7 × 10^-19 J

Writing in electron volts, we get,

⇒ V = -19.2 eV

(∵ 1eV = 1.6 × 10^-19 J)

Therefore, the potential energy of the system is -19.2 eV.

Let a be the radius of the sphere A, Q_A be the charge on the sphere, and C_A be the capacitance of the sphere. Let b be the radius of the sphere B, Q_B be the charge on the sphere, and C_B be the capacitance of the sphere. Since, the two spheres are connected with wire, their potential V is equal.

Electric field generated by a charge, E =

Where,

q = charge

d = distance from origin

ϵ₀ = permittivity of space

Let E_A be the electric field for sphere A and E_B be the electric field for sphere B. Therefore their ratio,

=

However, = and =

⇒ =

∴ = =

Hence, the ration of electric fields at surface is

The charges –q and + q at (0,0,-a) and (0,0,a) respectively forms a dipole.

(a) The point (0,0,z) lies on the axis of dipole, while the point (x,y,0) lies on the equatorial plane of the dipole.

The electric potential of dipole on any point on its axis is given by the formula–

where, = Absolute Permittivity of free space

p = Dipole moment of the system of two charges

r = Distance of point from the centre of the dipole

here, p = 2qa and r = z.

The electric potential of dipole at point (0,0,z) =

The electric potential of dipole on any equatorial point is zero

The electric potential of dipole at point (x,y,0) = 0.

(b) Given r/a>>1, which implies r>>a

the distance of point where potential is to be obtained is much greater than half of the distance between the two charges.

Hence, the potential (V) at a distance r is inversely proportional to square of the distance, i.e. V 1/r²

(c) A test charge is moved from point (5, 0, 0) to point (-7, 0, 0) along the x-axis.

Electrostatic potential (V1) at point (5, 0, 0) is given by,

V1 =

=

= 0

Electrostatic potential(V2) at point (-7, 0, 0) is given by,

V2 =

=

= 0

The potential difference between points (5,0,0) and (-7,0,0,) is V1-V2 = 0.

Thus,

The work done in moving a small test charge from the point (5,0,0) to (–7,0,0) along the x-axis = 0.

The answer does not change if the path of the test charge between the same points is not along the x-axis because work done by the electrostatic field in moving a test charge between the two points is independent of the path connecting the two points.

The diagram is given as:

X Y Z

The given charges of same magnitude placed at points X, Y, and Z respectively, forms an electric quadrupole.

Where, charge + q is at point X, charge -2q is at point Y, and charge + q is at point Z.

The point P is at a distance r from point Y.

Here, XY = YZ = a

So, YP = r, PX = r + a, PZ = r-a.

The electrostatic potential due to the system of three charges at point P is given by,

∴

⇒

⇒

⇒

⇒

⇒

Since, r/a>>1

Then a/r<<1 which implies a²/r² is negligible so it can be ignored,

Thus, potential becomes,

V =

Relation of potential due to quadrupole with r is V1/r³

However, for a dipole it is V1/r²

And for a monopole it is V1/r

Potential difference across the circuit = 1kV = 1000V

Capacitance of each capacitor = 1 μF

Potential difference each capacitor can withstand = 400V

Capacitance required across the circuit = 2 μF

Assume n number of capacitors are connected in series and further m number of such series circuits are connected in parallel to each other.

As the potential difference in the circuit is 1000V so the potential difference across each row of n capacitors is 1000V, as the potential difference each capacitor can withstand is 400V,

Therefore, 400V × n = 1000V

⇒ n = 1000V/400V = 2.5~3capacitors in each row.

Now,

Capacitance of each row = μF

As there are m column of three capacitors, which are connected in parallel.

Hence, equivalent capacitance of the circuit is

+ ……….m times =

Since, Required capacitance = 2 μF

⇒ m = 6

Total number of capacitors = 3 × 6 = 18.

Let the area of the plates of capacitor be A.

Capacitance of a parallel capacitor (C) = 2 F

Distance between the two plates (d) = 0.5cm = 0.005m

Capacitance of a parallel plate capacitor is given by the relation,

Where,

= Absolute Permittivity of free space = 8.8510^-12 C²N^-1m^-2

= 0.001129910¹²m² ~ 1130km²

As the area of the plates required becomes too large so the capacitance is taken in the range of μF .

Capacitance of capacitor C₁ = 100pF, C2 = 200pF, C3 = 200 pF, C4 = 100 pF.

Supply potential (V) = 300 V

Let the equivalent capacitance of capacitors C₂ and C₃(connected in series) be C'.

⇒ C’ = 100pF

Let the equivalent capacitance of capacitors C₁ and C’(connected in parallel) be C’’.

C’’ = C’ + C₁ = 100 + 100 = 200pF

Let the equivalent capacitance of C’’ and C4 (connected in series) be C’’’.

⇒ C’’’ = 200/3pF

Thus, the equivalent capacitance of the circuit = 200/3 pF.

Now, Supply potential (V) = 300 V

Charge on C₄ is given by,

Q₄ = C’’’V = = 2 × 10^-8C

So, V₄ = Q₄/C₄ = (210^-8)/(10010^-12) = 200C

Potential difference across C₁ is given by, V1 = V-V₄ = 300-200 = 100V

Charge on C₁ is given by, Q1 = C₁V₁ = 100 × 10^-12 × 100 = 10^-8C

C₂ and C₃ having same capacitances have a potential difference of 100 V together.

Since C₁ and C₃ are in series, the potential difference across C₂ and C₃ is given by,V₂ = V₃ = 50V

Therefore charge on C₂ is given by,Q₂ = C₂V₂ = 20010^-1250 = 10^-8C

And charge on C₃ is given by,Q₃ = C₃V₃ = 20010^-1250 = 10^-8C.

The charge and voltage across each capacitor are as follows:

Area of the plates of a parallel capacitor (A) = 90cm² = 910^-3m

Distance between the plates (d) = 2.5mm = 2.510^-3m

Potential difference across the plates (V) = 400V

(a) Electrostatic energy stored in the capacitor(E) = CV²

Since, Capacitance (C) =

∴

Where, = Absolute Permittivity of free space = 8.8510^-12C²N^-1m^-2

⇒ E = 2.5510^-6J

The electrostatic energy stored by the capacitor = 2.5510^-6J

(b) Volume between plates of capacitor (v) = A × d

= 910^-3m² × 2.510^-3m

= 2.2510^-5m³

Therefore, Energy per unit volume (u) = E/v

The energy per unit volume (u) = 0.113 Jm^-3

Relation between u and the magnitude of electric field E between the plates

Since,

Energy per unit volume (u) = E/v

= CV² /v

Here, V = voltage across capacitor

v = volume between plates of capacitor,

Here, V/d is the Electric intensity or Electric field so,

Capacitance of a charged capacitor(C) = 4 μF = 410^-6F

Voltage supplied to the capacitor (V) = 200 V

Electrostatic energy of the capacitor (E) = CV² = = 810^-2 J

Capacitance of an uncharged capacitor (C’) = 2 μF = 210^-6F

When both the capacitors are connected in a circuit, then initial charge on charged capacitor is equal to the final charge on both the capacitors in circuit. (According to law of conservation of charges)

Since, Charge = Voltage Capacitence

Therefore, CV = (C + C’) V’, where V’ is the voltage in the circuit when both capacitors are connected.

410^-6F200V = (410^-6F + 210^-6F) V’

⇒ V = V

Now, Electrostatic energy for the combination of two capacitors = 1/2 (C + C’)V’²

∴ U = 5.3310^-2 J

Thus, the electrostatic energy of the first capacitor is lost in the form of heat and electromagnetic radiation = 810^-2 J-5.3310^-2 J = 6.6710^-2J

Let the force applied to separate the plates of a parallel plate capacitor by a distance of x be F. Hence, work done = Fx.

The increase in potential energy of the capacitor = uv = uAx

Where, u = Energy density, A = area of each plate and v = volume between plates of capacitors.

Since, work done will be equal to the increase in the potential energy i.e.,

work done = increase in potential energy of the capacitor

Fx = uAx

F = uA

Since, u = and = E (Electric intensity or Electric field)

and

Therefore, F = 1/2 (CV)E and Q = CV

Thus, F = QE

The physical origin of the factor in the force formula lies in the fact that just outside the conductor, electric field is E and inside it is zero. Hence, it is the average value of the field that contributes to the force.

Radius of the outer shell = r₁

Radius of the inner shell = r₂

Charge on the inner surface of the outer shell = Q

Charge on the outer surface of the inner shell = -Q

Potential difference between the two shells,

Where, = Absolute Permittivity of free space = 8.8510^-12C²N^-1m^-2

Since, Capacitance,

Hence, proved.

Radius of the outer shell (r₁) = 13cm = 0.13m

Radius of the inner shell (r₂) = 12cm = 0.12m

Charge on the outer surface of the inner shell = 2.5 μC = 2.510^-6C

Dielectric constant of liquid () = 32

Since, Potential difference between the two shells,

∴ Δ

Where, = Absolute Permittivity of free space = 8.8510^-12C²N^-1m^-2

Therefore

(a) Capacitance,

⇒ C = 5.5510^-9F

Hence, the capacitance of the capacitor = 5.5510^-9F

(b) The potential of a capacitor (V), is given as:

⇒

⇒ V = 450 V

Hence, the potential of the inner sphere = 450V

(c) Radius of the isolated sphere(R) = 12 cm = 0.12m

Capacitance of an isolated sphere = 4πε₀R

∴ C = 4 × 3.14 × 8.85 × 10^-12 × 0.12

C = 1.3310^-11F

Where, ϵ₀ is the Absolute Permittivity of free space = 8.8510^-12C²N^-1m^-2The capacitance of the isolated sphere is less than that of the concentric spheres because the outer sphere of the concentric spheres is earthed. Hence, the potential difference is less and the capacitance is more than the isolated sphere.

(a) According to question, two large conducting spheres with charge Q₁ and Q₂ are placed such that distance between their centre is r. Then, the magnitude of electrostatic force between the two spheres is not exactly equal to

Since, the magnitude of electrostatic force between two-point charges Q₁ and Q₂ placed at a distance r is given by the relation,

.

Here both the spheres have non-uniform distribution of charges Q₁ and Q₂ on their surface.

(b) Gauss’s law states that net electric flux through a surface is the electric field of the charge enclosed by the surface multiplied by the area of the surface projected in a plane perpendicular to the field and this product is a constant equal to charge enclosed within the surface divided by ϵ₀

i.e.

or

Where, electric flux,

E = electric field

and dA is the elementary area vector of the surface, Q = charge enclosed by the surface, ϵ₀ = permittivity of free space.

Since, Electric field involves 1/r² and area vector involves the term of r², therefore their product becomes constant.

But, If Coulomb’s law involved 1/r³ dependence (instead of 1/r²), product of electric field and area vector Gauss’s law will not be true.

(c) If a small test charge is released at rest at a point in an electrostatic field configuration, then it will travel along the field lines passing through that point, as the tangent drawn at a point on the field line gives the direction of acceleration at that point.

(d) When the electron completes an orbit, either circular or elliptical, the displacement becomes zero, so the work done by the field of a nucleus is zero. ∵ Work = force × displacement

(e) Electric field is discontinuous across the surface of a charged conductor, but the electric potential is continuous.

(f) The capacitance of a single conductor means a capacitance of a parallel plate capacitor with one plate as the conductor while the other placed at infinity.

(g) Water has a permanent dipole moment while mica doesn’t, because water has an unsymmetrical space as compared to mica. So, water has a greater dielectric constant than mica.

Radius of outer cylinder(R) = 1.5cm = 0.015m

Radius of inner cylinder(r) = 1.4cm = 0.014m

Charge on the inner cylinder(q) = 3.5m = 3.5 × 10^-6C

Length of a co-axial cylinder(l) = 15cm = 0.15m

Capacitance of a co-axial cylinder of radii R and r, is given as:

where ϵ₀ = Permittivity of free space = 8.8510^-12N^-1 m^-2 C²

Therefore,

(Trick: Multiply by 2.303 to convert log_e to log₁₀)

∴ C = 1.210^-10F

Potential difference of the inner cylinder,

∴

∴ V = 2.2910⁴V

Therefore, Capacitance = 1.210^-10F

Potential difference = 2.2910⁴V

Potential difference of a parallel plate capacitor(V) = 1kV = 1000V

Dielectric constant of a material(ϵ_r) = 3

Dielectric strength = 10⁷V/m

Electric field intensity(E) = 10%of 10⁷

⇒ E = 10⁶V/m

(since, the field intensity never exceeds 10% of the dielectric strength)

Capacitance of the parallel plate capacitor(C) = 50pF = 50 × 10^-12F

Distance between the plates (d),

⇒

⇒ d = 10^-3m

∵ Capacitance =

where, ϵ₀ = Permittivity of free space = 8.85 × 10^-12N^-1 m^-2 C²

A = area of plates

∴

⇒ A = 50 × 10^-12F × 10^-3m/8.85 × 10^-12N^-1 m^-2 C² × 3

⇒ A = 1.88 × 10^-3m²

∴ Minimum area of the plates required = 1.88 × 10^-3m²

(a) The equipotential surfaces for a constant electric field in the z-direction are the equidistant planes parallel to the x-y plane.

(b) The equipotential surfaces for a field that uniformly increases in magnitude but remains in a constant in z direction are the planes parallel to the x-y plane and this field increases when the planes get closer.

For both cases A and B the Equipotential Surface is same as below

(c) The equipotential surfaces for a single positive charge at the origin are the concentric spheres centered at the origin.

D. The equipotential surface for a uniform grid consisting of long equally spaced parallel charged wires in a plane, near the grid is shaped as a periodically alternating curve. At a larger distance from the grid, the surface becomes parallel to the grid.

Potential Difference of the spherical metal shell (V) = 15 × 10⁶ V

Dielectric strength of the gas surrounding the electrode = 5 × 10⁷Vm^-1

Dielectric strength = Electric field intensity (E)

The minimum radius of the spherical shell (r),

∴ r = 15 × 10⁶ V/5 × 10⁷ Vm^-1 = 0.3m

According to Gauss’s law, the electric field between a sphere and a shell is determined by the charge on the smaller sphere.

∴ The potential difference (V) between the sphere and the shell depends on q₁ and is independent of charge q₂.

∵ q₁ is positive so V is positive which implies charge will necessarily flow from the sphere to the shell.

(a) We do not get an electric shock as we step out of our house because the original equipotential surfaces of open air changes, and our body being a good conductor suitably comes in equipotential with the earth surface.

(b) Yes, the man will get an electric shock if he comes in contact with the metal slab next morning. The steady discharging current in the atmosphere charges up the aluminium sheet. As a result a substantial potential is built gradually. The raise in the voltage depends on the capacitance of the capacitor formed by the aluminium slab and the ground.

(c) The occurrence of thunderstorms and lightning charges the atmosphere continuously. So, during lightning even with the presence of a discharging current of 1800 A, the atmosphere is not discharged completely. The two opposing currents are in equilibrium and the atmosphere remains neutralized.

(d) During lightning and thunderstorm, heat energy, light energy and sound energy are dissipated in the atmosphere.