Prove that total electrostatic potential energy stored in a capacitor is 1 2 CV 2 . Hence, derive an expression for the energy density of an electric field.

Expression for the stored energy- Suppose a parallel plate capacitor has capacitance.On plats charges are + q and - q and their potential differences is V . ∴ V = q / C . . . . . . . . . . . . ( i ) Now d q charge is transferred than work done. d w = v d q = q / C d q from(i) Now work done in transferring total Q charge. This work is stored in capacitor as potential energy ( U ) . Expression for energy density-

Prove that total electrostatic potential energy stored in a capacitor is 1 2 CV 2 . Hence, derive an expression for the energy density of an electric field.

Expression for the stored energy- Suppose a parallel plate capacitor has capacitance.On plats charges are + q and - q and their potential differences is V . ∴ V = q / C . . . . . . . . . . . . ( i ) Now d q charge is transferred than work done. d w = v d q = q / C d q from(i) Now work done in transferring total Q charge. This work is stored in capacitor as potential energy ( U ) . Expression for energy density-

Three capacitors are connected as shown in the following figure. Calculate the resultant capacitance. If the combination is connected across a 30 V DC battery, find the total charge drawn by the arrangement.

V = 30 v o l t . C 2 a n d C 3 a r e i n s e r i e s c o m b i n a t i o n 1 C ‵ = 1 2 + 1 3 = 3 + 2 6 = 5 6 . C ' = 6 5 M F . C 1 a n d C ' a r e i n p a r a l l e l c o m b i n a t i o n . C = C 1 + C ' C = 1 + 6 5 , C = 5 + 6 5 = 11 5 M F . C = 11 5 M F . Q = C V , Q = 11 5 × 30 . Q = 66 M C .

Two charges of + 5 nC and - 2 nC are placed at points ( 5 cm , 0 , 0 ) and ( 23 cm , 0 , 0 ) in a region of space where there is no other external electric field. Calculate potential energy of this charge system.

Electrostatic potential energy of the system of charge is U = 1 4 πε o q 1 q 2 r = 9 × 10 9 × 5 × 10 - 9 × - 2 × 10 - 1 0 . 18 = - 9 × 5 × 2 × 10 - 9 × 10 2 18 U = - 5 × 10 - 7 J ..

Two point charges of + 1 . 5 μC and + 2 . 5 μC are placed 30 cm apart. Calculate the magnitude of electric potential and electric field at the midpoint of the line joining the two charges.

Two charges placed at points A and B are represented in the given figure. O is the mid-point of the line joining the two charges. Magnitude of charge located at A, q 1 = 1 . 5 μ C Magnitude of charge located at B, q 2 = 2 . 5 μ C Distance between the two charges, d = 30 c m = 0 . 3 m (a) Let V 1 and E 1 are the electric potential and electric field respectively at O. V 1 = Potential due to charge at A + Potential due to charge at B. V 1 q 1 4 πε o d 2 + q 2 4 πε o d 2 = 1 4 πε o d 2 q 1 + q 2 ε o = Permittivity of free space 1 4 πε o = 9 × 10 9 N C 2 m - 2 ∴ V 1 = 9 × 10 9 × 10 - 6 0 . 30 2 E 1 = Electric field due to q 2 - Electric field due to q 1 . = q 2 4 πε o d 2 2 - q 1 4 πε o d 2 2 = 9 × 10 9 0 . 30 2 2 × 10 6 × 2 . 5 - 1 . 5 = 4 × 10 5 V m - 1 Therefore, the potential at mid-point is 2 . 4 × 105 V and the electric field at mid-point is 4 × 105 V m - 1 . The field is directed from the larger charge to the smaller charge. ..

Establish a relationship between electric potential and electric field.

Consider two point A and B separated by a small distance d x is small, the electric field E is assumed to be uniform along A B .The force acting on a unit positive charge at A is equal to E . Now, the work done in moving a unit positive charge from A to B against the electric field is d W = - E × d x . The negative sign shows that the work is done against the direction of the field.Since the work done is equal to potential difference dV between A and B then d V = - E × d x o r E = - d V d x . Thus, the electric field at a point is the negative of the potential gradient at that point.

E M B I B E

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR PHYSICS>Chapter 2 - Electrostatic Potential and Capacitance>Nagaland Board-2019>Q 5

HARD

12th Nagaland Board

IMPORTANT

Earn 100

Prove that total electrostatic potential energy stored in a capacitor is $\frac{1}{2} {CV}^{2}$ . Hence, derive an expression for the energy density of an electric field.

Important Questions on Electrostatic Potential and Capacitance

EASY

12th Nagaland Board

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR PHYSICS>Chapter 2 - Electrostatic Potential and Capacitance>Nagaland Board-2020>Q 6

Derive an expression for electric potential due to a point charge. How do electric potential vary with distance

‘ r ’

for a point charge?

EASY

12th Nagaland Board

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR PHYSICS>Chapter 2 - Electrostatic Potential and Capacitance>Nagaland Board-2020>Q 7

A parallel plate capacitor with air between the plates has a capacitance of

8 pF (1 pF = 10^{- 12} F)

. What will be the capacitance of the distance between the plates is reduced by half, and the space between them is filled with a substance of dielectric constant