Three-point charges q , – 4 q and 2 q are placed at the vertices of an equilateral triangle ABC of side l as shown in the figure. Find out the amount of the work done to separate the charges at infinite distance.

The work done to separate the charges to infinity = Change in the potential energy of the system ⇒ W = ∆ P . E . = U f - U i The initial potential energy of the system, U i = 1 4 πε 0 ( - 4 q ) ( 2 q ) l + ( 2 q ) ( q ) l + ( - 4 q ) ( q ) l U i = q 2 4 πε 0 - 8 + 2 - 4 l = - 10 q 2 4 πε 0 l Final potential energy, U f = 0 Work done, W = U f - U i = 0 - - 10 q 2 4 πε 0 l = 10 q 2 4 πε 0 l Thus, W = 5 q 2 2 πε 0 l

The space between the plates of a parallel plate capacitor is completely filled in two ways. In the first case, it is filled with a slab of dielectric constant K . In the second case, it is filled with two slabs of equal thickness and dielectric constants K 1 and K 2 respectively as shown in the figure. The capacitance of the capacitor is same in the two cases. Obtain the relationship between K , K 1 and K 2 .

In case I : Dielectric constant is K , area of the cross-section is A and width of dielectric slab is d . Then, the capacitance will be: C I = K ε ∘ A d In case II : For first d 2 space is covered by dielectric slab K 1 with cross-sectional area A and the next d 2 space is covered by dielectric slab K 2 with cross-sectional area A . Then the total capacitance here is: 1 C II = 1 C K 1 + 1 C K 2 1 C II = d 2 K 1 ε ∘ A + d 2 K 2 ε ∘ A = d 2 ε ∘ A K 1 + K 2 K 1 K 2 ⇒ C II = 2 ε ∘ A d K 1 K 2 K 1 + K 2 Since, both the capacitance are equal. Then, C I = C II K ε ∘ A d = 2 ε ∘ A d K 1 K 2 K 1 + K 2 ⇒ K = 2 K 1 K 2 K 1 + K 2 Hence, the relation between K , K 1 and K 2 is K = 2 K 1 K 2 K 1 + K 2 .

A parallel plate capacitor is charged by a battery to a potential difference V. It is disconnected from battery and then connected to another uncharged capacitor of the same capacitance. Calculate the ratio of the energy stored in the combination to the initial energy on the single capacitor.

We know, the energy stored by the capacitor when it is charged is given as : U i = 1 2 Q 2 C = 1 2 C V 2 Now, when the capacitor is connected to the other Capacitor, they both get some charge in equilibrium. Hence, the total charge remains the same, the charge on each capacitor is : Q f = Q 2 Now, the energy stored is - U f = 2 × 1 2 Q f 2 C = 1 4 Q 2 C = 1 4 C V 2 Hence, the Ratio is : U f U i = 1 2

Describe briefly the process of transferring the charge between the two plates of a parallel plate capacitor when connected to a battery. Derive an expression for the energy stored in a capacitor.

Let us consider, an uncharged capacitor of capacitance 'C' which is connected with a battery of e.m.f. Potential Difference across the capacitor will build up. Such that, the plate with the positive terminal of the battery will lose electrons and the other plate will gain the electron. There is no transfer of charge in between the plates. When the emf of the battery becomes equal to the potential difference across the capacitor, then the capacitor would be fully charged and will act as an open switch. Let the charge on the capacitor is increased by d q amount then, Then, the work done : W = d U = V d q d U = q C d q Total energy needed to increase the charge from zero to Q U = ∫ 0 Q q d q C U = 1 2 Q 2 C = 1 2 C V 2

Four-point charges Q , q , Q and q are placed at the corners of a square of side ' a ' as shown in the figure. Find the potential energy of this system.

The potential energy between two charges q 1 and q 2 separated by a distance can be given as U = 1 4 πε 0 q 1 q 2 r From the diagram, the adjacent charges are kept at equal distances. Thus, by symmetry, the system potential energy between charges that are adjacent to each other will be the same, and it can be given as: U 1 = 4 × 1 4 πε 0 q Q a For the diagonally opposite charge q , the Potential energy will be: U 2 = 1 4 πε 0 q 2 a 2 For the diagonally opposite charge Q , the potential energy will be: U 3 = 1 4 πε 0 Q 2 a 2 Thus, the total potential energy of the system, U = U 1 + U 2 + U 3 U = 4 × 1 4 πε 0 q Q a + 1 4 πε 0 q 2 a 2 + 1 4 πε 0 Q 2 a 2 U = 1 4 πε 0 a 4 q Q + q 2 2 + Q 2 2

Derive an expression for the electric potential at any point along the axial line of an electric dipole.

Consider a dipole ± q having length 2 l . Let a point P be at a distance of r from the Centre of dipole. Electric dipole moment, p = q . 2 l , directed from - q to + q . The potential at P due to + q is V 1 = 1 4 π ε ∘ · q r - l The potential at P due to - q is V 2 = - 1 4 π ε ∘ · q r + l Thus, resultant potential at P due to dipole V = V 1 + V 2 V = 1 4 π ε ∘ · q r - l - 1 4 π ε ∘ · q r + l ⇒ V = 1 4 π ε ∘ · q 1 r - l - 1 r + l ⇒ V = 1 4 π ε ∘ · q 2 l r 2 - l 2 or, V = 1 4 π ε ∘ · p r 2 - l 2 If point P is far away from the dipole, r > > l , then V = 1 4 π ε 0 p r 2

Draw the equipotential surfaces corresponding to a uniform electric field in the z ‐ direction.

As the uniform electric field is in the Z-direction, the equipotential surfaces are normal to the field, i.e. in XY-plane. The equipotential surfaces are equidistant from each other. Hence, the equipotential surfaces corresponding to a constant electric field in the z -direction are the respective planes that are parallel to X - Y plane. Each of these planes is an equipotential surface having its unique value of field potential, and the electric field is always perpendicular to these planes.

E M B I B E

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR PHYSICS>Chapter 2 - Electrostatic Potential and Capacitance>cbse-2020>Q 9

EASY

12th CBSE

IMPORTANT

Earn 100

Draw equipotential surface due to an isolated point charge $(- q)$ and depict the electric field lines.

Important Questions on Electrostatic Potential and Capacitance

MEDIUM

12th CBSE

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR PHYSICS>Chapter 2 - Electrostatic Potential and Capacitance>cbse-2020>Q 10

Three point charges $+ 1 μ C, - 1 μ C$ and $+ 2 μ C$ are initially infinite distance apart. Calculate the work done in assembling these charges at the vertices of an equilateral triangle of side $10 cm$ . Take $(\frac{1}{4 π ε_{o}}) = 9 \times 10^{9} N m^{2} C^{- 2}$