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i. When a petrol or a diesel tanker is emptied in a tank, it is grounded so that it has an electrically conductive connection from the petrol or diesel tank to ground (Earth) to allow leakage of static and electrical charges. 

ii. Metallic bodies of cars, trucks or any other big vehicles get charged because of friction between them and the air rushing past them. Hence, a thick chain is hanged from a petrol or a diesel tanker to make a contact with ground so that charge produced can leak to the ground through chain.

The electric potential at a point P is equal to the work done by an external force to bring a unit positive charge with constant velocity from infinity to the point P in the region of the external electric field \(\vec E\)

Correct answer is (c) pE

correct answer is (c) p and r^-3

(a) pE (1 – cos θ)

W = pE(cos θ₀ – cos θ) 

[θ₀ = cos 0, cos 0 = 1] 

W = pE(1 – cos θ)

The electric dipole moment vector lies along the line joining two charges and is directed from -q to + q. The SI unit of dipole moment is coulomb meter (Cm).

\(\vec P\) = qa \(\hat i\) -qa (- \(\hat i\)) = 2qa \(\hat i\) 

Correct option is: (B) pE sinθ

i. Similarities:

a. Both forces obey inverse square law:

F ∝ \(\frac{1}{r^2}\)

b. Both are central forces and they act along the line joining the two objects.

ii. Differences:

a. Gravitational force between two objects is always attractive while electrostatic force between two charges can be either attractive or repulsive depending on the nature of charges.

b. Gravitational force is about 36 orders of magnitude weaker than the electrostatic force.

U = \(\cfrac12\)QV = \(\cfrac12\) x 2 x 10^-6 C x 100V

= 10^-4 J

is the required energy.

Yes. To charge a capacitor, an external agent (the battery) has to do work against the electrostatic force due to the charges already present on the plates of the capacitor. This work done is stored in the form of potential energy in the electric field in the medium between the plates of the capacitor.

(i) If the plates of a charged capacitor are moved farther apart after the battery is disconnected, the energy stored increases by the amount of work done by the external agent in pulling the plates apart against the force of attraction between the opposite charges on the plates.

(ii) With the battery still connected, increasing the separation between the plates decreases the energy stored in the charged capacitor.

[Notes : (1) The charge on the capacitor does not change after the battery is disconnected. Because the electric field of a large plate is independent of the distance from the plate, the electric field between the plates also remains the same. Therefore, since E = V/d, the p.d. between the plates (V) changes in the same proportion as d. Since the energy stored,

U = \(\cfrac 12CV^2\) = \(\cfrac12\)\(\left(\cfrac{A_ɛ}d \right)\)(Ed)² = \(\cfrac12\) (AɛD)E²

U also changes in the same proportion as d. The addi-tional energy is transferred to the system from the work done by the external agent.

(2) With the battery still connected, the p.d. between the plates remains the same. Since the energy stored,

U = \(\cfrac 12CV^2\) = \(\cfrac12\)\(\left(\cfrac{A_ɛ}d \right)\)V².

U is inversely proportional to d. With decrease in capacitance, the charge on the plates decreases.]

The correct answer is option (d)

Explanation:

V_i = 50 v
V_f = 40 v 
we know that fraction of stored energy is given by 
[1/2 C(V_i)²] / [1/2 C(Vf)²]

= (40/50)² = 16/25

= 0.64

Data: p = 6.3 × 10^-30 C∙m, N = 10²¹ molecules, 

E = 2.5 × 10⁵ N/C, θ₀ = θ₁ = 0°, θ = θ₂ = 90° 

W = pE(cos θ₀ – cos θ)

The total work required to orient N dipoles is 

W = NpE(cos θ₁ – cos θ₂)

=(10²¹)(6.3 × 10^-30)(2.5 × 10⁵) 

= 15.75 × 10^-4 J = 1.575 mJ

The electrostatic potential at each point in space in the vicinity of the source charges represents a scalar field.

The advantages of electrostatic potential field associated with a given distribution of charges are as follows:

• If we know the potential difference between any two points, we can easily obtain the change in potential energy and the work done when a charge placed in the field moves between these two points.
• Electric field is a vector field. Electrostatic potential being a scalar field, the potential at any point due to several charges is simply the algebraic sum of the potentials due to the individual charges.
• The construction of equipotential surfaces helps to visualize the electric field pattern.
• It is possible to calculate the electric field \(\vec E\) from the scalar potential field function V (by differentiating V with respect to the space coordinates.)

V = \(\cfrac1{4πε_0}\) , \(\cfrac Qr\)

= \(\cfrac{9\times10^{9}N.m^2/C^2\times10^{-18}C}{10\times10^{-10}m}\)

= 9 volts is the required electric potential.

Definition The rate of change of electric potential with distance in a specified direction is called the electric potential gradient in that direction.

SI unit of energy : the joule (J) 

Non-SI unit of energy : the electronvolt (eV).

In electrostatics, one joule is the change in electric potential energy when a charge of one coulomb is moved through a potential difference of 1 volt. Therefore, 1J = 1C × 1V, so that 1V = 1J/C.

A charged spherical conductor is equivalent to a point charge at its centre. For U(∞) = 0, the potential energy of two point charges q₁ and q₂ a distance r apart is

U = \(\cfrac1{4πε_0}\)\(\cfrac{q_1q_2}{r_{21}}\)

Since the FE of the charged shell and electron system is negative, the shell must be positively charged, electron being negatively charged.

If the electron is replaced by a proton, the PE of the new system would be positive equal to 

U’ = ±15 × 10^-20 J.

Consider a charge q placed in an external electric field at a point whose position vector with respect to an arbitrary reference frame is \(\vec r\). If V (\(\vec r\)) is the potential of the point, with respect to an arbitrary reference zero at infinity, then the potential energy of the charge q at the point is

U (\(\vec r\)) = qV(\(\vec r\))

where it is assumed that q is sufficiently small and does not significantly distort the electric field and the potential at the point.

A charge in an electric field possesses electric potential energy just as a particle in a gravitational field possesses gravitational potential energy. Consider a test charge q₀ in an electric field, moved very slowly by an external agent from point B where its electric potential energy is U_B to a point A where its electric potential energy is U_A.

The change in the potential energy, U_A– U_B is defined as the work W_B→A that must be done by an external agent to move the test charge from B to A against the electric force, keeping the charge always in equilibrium, i.-e., without accelerating the charge so as not to give it any kinetic energy.

W_B→A = ∆U = U_A – U_B

The potential difference ∆V = V_AB = V_A - V_B between two points A and B in electric field is ∆V = \(\cfrac{W_{B→A}}{q_0}\) = \(\cfrac{ΔU}{q_0}\)

Definition : The electric potential difference between two points in an electric field is defined as the work done per unit charge by an external agent against the electric force in moving an infinitesimal positive charge from one point to the other without acceleration.

We choose the potential energy U_B and potential V to be zero when the initial point B is infinitely far from the source charges which produce the field. Then, the work done per unit test charge by an external agent in bringing a test charge from infinity to a point is the electric potential at that point. The electric potential at a distance r from a source charge,

V(r) = \(\cfrac{W_{∞→r}}{q_0}\) = \(\cfrac{U(r)}{q_0}\)

Definition : The electric potential V at a point in an electric field is defined as the work per unit charge that must be done by an external agent against the electric force to move without acceleration a sufficiently small positive test charge from infinity to the point of interest.

[Note : It is more correct to speak about potential difference ∆V between two points than just the potential V at a given point because the latter implies a choice of .zero reference potential. The choice is one of convenience and we may choose the zero reference potential for a point at infinity or at some other location convenient for the problem as is done for gravitational potential.]