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-

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 4

EASY

12th Nagaland Board

IMPORTANT

Earn 100

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.

Important Questions on Electrostatic Potential and Capacitance

HARD

12th Nagaland Board

IMPORTANT

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

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.

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