Derive an expression for the electric field at any point on the equatorial line of an electric dipole.

Consider an electric dipole consisting of charges - q and + q separated by a distance 2 a . Let E A be the electric field at point Q due to charge - q and E B be the electric field at point Q due to charge + q where Q is a point on the equatorial line of the dipole whose electric field is to be measured. Let x be the distance between the centre of the dipole and point Q . Now, resultant electric field at point Q would be, E Q = E A + E B E A = 1 4 π ε 0 q A Q 2 ⇒ 1 4 π ε 0 q x 2 + a 2 ( ∵ According to Pythagoras, A Q = x 2 + a 2 2 ) E B = 1 4 π ε 0 q B Q 2 ⇒ 1 4 π ε 0 q x 2 + a 2 ( ∵ According to Pythagoras, B P = x 2 + a 2 2 ) Now, according to parallelogram law of vector addition, the resultant electric field would become, E Q = E A 2 + E B 2 + 2 E A E B cos 2 θ Now, since E A = E B we can say, E A = E B = E ' E Q = E ' 2 + E ' 2 + 2 E ' E ' cos 2 θ ⇒ 2 E ' 2 ( 1 + cos 2 θ ) ⇒ E ' 2 ( 1 + cos 2 θ ) ⇒ E ' 2 2 cos 2 θ E Q = 2 E ' cos θ - - - ( i ) Now, from △ A O Q , cos θ = a x 2 + a 2 Also, we know that, E ' = 1 4 π ε 0 q x 2 + a 2 Putting both these values in equation ( i ) , we get, E Q = 2 4 π ε 0 q x 2 + a 2 a x 2 + a 2 E Q = p 4 π ε 0 x 2 + a 2 3 2 Since we know that in dipoles, x > > > 2 a , then a 2 can be neglected. ∴ E = p 4 π ε 0 x 3 2

The identical point charges, q each, are kept 2 m apart in air. A third point charge Q of unknown magnitude and sign is placed on the line joining the charges such that the system remains in equilibrium. Find the position and nature of Q .

= Let the two charges of + q each placed at point A and B at a distance 2 m apart in air. Suppose, the third charge Q (unknown magnitude and charge) is placed at a point O, on the line joining the other two charges, such that O A = x , O B = 2 - x . For the system to be in equilibrium, the net force on each charge must be zero. If we assume that charge Q placed at O is positive, the force on it may be zero. But the force on charge q at point A or B will not be zero. It is because , the forces on a charge q due to the other two charges (Q and q) will act in the same direction. If charge Q is negative, then the forces on vharge q due to the other two charges will act in opposite direction. Hence, Q will be negative in nature. For the charge ( - Q ) to be in equilibrium Force on charge ( - Q ) due to charge ( + q ) at point A should be equal and opposite to charge ( + q ) at B 1 4 πε 0 Q q x 2 = 1 4 πε 0 Q q 2 - x 2 On solving this we get : x = 1 m

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. Obtain the expression for the magnitude of the resultant electric force acting on the charge q .

The forces acting on the charge q are due to the other two charges, The force at A due to charge at C, F 1 = k q 2 q l 2 = 2 k q 2 l 2 The force at A due to charge at B, F 2 = k q - 4 q l 2 = - 4 k q 2 l 2 The forces acting on charge q can be broken into components as shown in the figure below: Resolving the forces along the x and y- directions. We can calculate the net force along the x-axis: F x = - F 1 cos 60 ° + F 2 cos 60 ° x ^ F x = - k 2 q 2 l 2 . 1 2 - k 4 q 2 l 2 . 1 2 x ^ = - 3 k q 2 l 2 x ^ The net force along the y-axis: F y = - F 2 sin 60 ° - F 1 sin 60 ° y ^ F y = - k 4 q 2 l 2 . 3 2 + k 2 q 2 l 2 . 3 2 x ^ = - 3 k q 2 l 2 y ^ Thus, the net force acting on the charge, F n e t = F x + F y F n e t = - 3 k q 2 l 2 x ^ - 3 k q 2 l 2 y ^ = - k q 2 l 2 3 x ^ + 3 y ^

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 resultant electric force on a charge Q .

According to the question, From the diagram given above, we can see, there are three forces due to the three other charges are acting on the charge Q kept at corner at one of the corners of the square. Applying the coulomb's law, the force acting between two charges q 1 and q 2 , kept at a distance r can be given as: F = k q 1 q 2 r 2 Using the coulomb's law to calculate forces acting between the charges kept at B and C, we get: F 1 = 1 4 πε 0 × q Q a 2 (along BC) Using the coulomb's law to calculate forces acting between the charges kept at D and C, we get: F 2 = 1 4 πε 0 × q Q a 2 (along DC) Using the coulomb's law to calculate forces acting between the charges kept at A and C, we get: F 3 = 1 4 πε 0 × Q 2 a 2 2 From the diagram, it can be seen that the forces F 1 and F 2 are perpendicular to each other. Thus, using the principle of superposition, the resultant force due to these two forces will be: F = F 1 2 + F 2 2 + 2 F 1 F 2 cos 90 ° Substitute, cos 90 ° = 0 F 1 = F 2 , we get: F = 2 F 1 2 = F 1 2 F = 2 4 πε 0 × q Q a 2 (Along AC) Thus, the net force on the charge at D can be calculated using the superposition principle, we get: F ' = F + F 3 F ' = 2 4 πε 0 × q Q a 2 + 1 4 πε 0 × q Q ( a 2 ) 2 F ' = Q 4 πε 0 2 q a 2 + Q 2 a 2 F ' = Q 4 πε 0 2 2 q + Q a 2 (along AC)

If the point charge is moved to a distance d from the centre of the square and the side of the square is doubled, explain how the electric flux will be affected.

On moving the charge to distance d from the centre of the square and making side of square 2 d , does not change the flux at all because the flux is independent of the side of the square or distance of charge in this case.

E M B I B E

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR PHYSICS>Chapter 1 - Electric Charges and Fields>cbse-2020>Q 11

EASY

12th CBSE

IMPORTANT

Earn 100

If the net electric flux through a closed surface is zero, then we can infer

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Important Questions on Electric Charges and Fields

EASY

12th CBSE

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR PHYSICS>Chapter 1 - Electric Charges and Fields>cbse-2020>Q 12

An electric dipole consisting of charges

+ q

and

- q

separated by a distance

L

is in stable equilibrium in a uniform electric field

\vec{E}

.The electrostatic potential energy of the dipole is :

EASY

12th CBSE

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR PHYSICS>Chapter 1 - Electric Charges and Fields>cbse-2020>Q 13

Using Gauss law, derive expression for electric field due to a spherical shell of uniform charge distribution $σ$ and radius $R$ at a point lying at a distance $x$ from the centre of shell, such that $0 < x < R$ .

MEDIUM

12th CBSE

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR PHYSICS>Chapter 1 - Electric Charges and Fields>cbse-2020>Q 14

Using Gauss law, derive expression for electric field due to a spherical shell of uniform charge distribution $σ$ and radius $R$ at a point lying at a distance $x$ from the centre of shell, such that $x > R$ .

MEDIUM

12th CBSE

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR PHYSICS>Chapter 1 - Electric Charges and Fields>cbse-2020>Q 15

An electric field is uniform and acts along $+ x$ direction in the region of positive $x$ . It is also uniform with the same magnitude but acts in $- x$ direction in the region of negative $x$ . The value of the field is $E = 200 N / C$ for $x > 0$ and $E = - 200 N / C$ for $x < 0$ . A right circular cylinder of length $20 cm$ and radius $5 cm$ has its centre at the origin and its axis along the $x -$ axis so that one flat face is at $x = + 10 cm$ and the other is at $x = - 10 cm$ . Find the net outward flux through the cylinder.

MEDIUM

12th CBSE

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR PHYSICS>Chapter 1 - Electric Charges and Fields>cbse-2020>Q 16

An electric field is uniform and acts along $+ x$ direction in the region of positive $x$ . It is also uniform with the same magnitude but acts in $- x$ direction in the region of negative $x$ . The value of the field is $E = 200 N / C$ for $x > 0$ and $E = - 200 N / C$ for $x < 0$ . A right circular cylinder of length $20 cm$ and radius $5 cm$ has its centre at the origin and its axis along the $x -$ axis so that one flat face is at $x = + 10 cm$ and the other is at $x = - 10 cm$ . Find the net charge present inside the cylinder. Take permittivity of free space $(ε_{o}) = 8.854 \times 10^{- 12} C^{2} N^{- 1} m^{- 2}$

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